Chapter 10: Electrochemistry

Electrochemistry Definition: The branch of chemistry which deals with the interconversion of electrical and chemical energy is called electrochemistry. OXIDATION, REDUCTION, AND REDOX REACTIONS Oxidation Oxidation is a process involving loss of electron or electrons. It is also the gain of oxygen and loss of hydrogen. But, here oxidation in terms of loss of electrons is discussed. Examples Fet? → Fet + e Zn →Zn2+ +ē Here Fet2 and Zn® have lost electrons, therefore they are oxidized. Reduction Reduction is a process involving gain of electron or electrons. It is also the gain of hydrogen and loss of oxygen. But, here reduction in terms of gain of electrons is discussed. Examples: Cl° +é → cl Cu2+ + 2é → Củ° In these examples, Cl° and Cut2 have gained electrons; therefore, they are reduced. Oxidation and reduction always take place together. The reaction in which this happens is called redox reactions. Redox Reaction: The reaction in which oxidation and reduction process takes place are called redox reactions. Example: The example of redox reaction is given below. Zn° + Cu2+ → Củ° + Znt2 In this reaction zinc undergoes oxidation as it lose two electrons as it goes to zero oxidation state from +2. Copper undergo reduction as it gets two electron and move from +2 to zero: oxidation state. Identify oxidation and reduction in a chemical reaction There are two ways of finding out whether or not a substance has been oxidized or reduced during a chemical reaction: • Electron transfer • Changes in oxidation number Interesting Information! S.Q. Write nature of photosynthesis and respiration reactions. Ans. Photosynthesis is a redox reaction which provides food for the entire planet, and another one is respiration that keeps us alive, both are redox reactions. OXIDATION NUMBER AND ITS SIGNIFICANCE

Oxidation number and its significance Oxidation number Definition: It is the apparent charge on an atom (per atom) of an element in a molecule or an ion. • The oxidation number is also called oxidation state. • Oxidation number is the measure of gain or lose electrons by an element. Significance of Oxidation number • An oxidation number is a number given to each atom or ion in a compound that show its degree of oxidation. • Oxidation number can be positive (+), negative (→) and zero (0). • The 't' or '-' sign must always be included when writing oxidation numbers. • A higher positive oxidation number means the atom or ion is more oxidized. • A higher negative oxidation number means the atom or ion is more reduced. OXIDATION NUMBER RULES Oxidation number rules The oxidation number of any atom or ion can be deduced by using oxidation number rules. It is important to note that an oxidation number refers to a single atom in a compound. Rules for assigning oxidation numbers Uncombined elements (i) The oxidation number of any uncombined (free) element is always zero. Examples: Oxidation Number of S8 ,Hz, Clz ,Na, Mg, Zn (in elemental form) is always zero (0). (i) Electronegativity rule In a compound or ion, the more electronegative element is given a negative oxidation number. Example: In HF, fluorine (more electronegative) gets -1, hydrogen gets +1. (ill) Fixed oxidation numbers for specific groups In compounds, many atoms or ions have fixed oxidation numbers. Group 1 (alkali metals) show always +1. Group2 (alkaline earth metals) show always +2. Group 17 (halogens) show -1 in binary compounds. Hydrogen usually show +1, but -1 in metal hydrides (NaH). Oxygen usually show -2, but in peroxides (H20z) -1 and in FO oxygen show +2 oxidation number. (iv) Monoatomic ions The oxidation number is equal to the charge on the ion in mono-atomic ions. Examples: Cl show oxidation number of -1. Al3* show oxidation number of +3. (v) Sum of oxidation numbers in compounds In a neutral compound, the sum of all oxidation numbers or algebraic sum is always zero. Examples: In HCl molecule H show +1, Cl show -1 and the sum is 0. In NaCl the Na show +1, Cl show -1. The sum of oxidation numbers is 0. (vi) Sum of oxidation numbers in ions In a polyatomic ion, the algebraic sum of oxidation numbers equals the charge of ion. , the sum of oxidation numbers is -2 Examples: In CO3" In SO* , the sum of oxidation numbers is -2 (vii) Variable oxidation states For the transition metals and many non-metals, the oxidation numbers is not fixed, because they show variable oxidation numbers. Their oxidation number are always find by calculations. Example: The oxidation number of Mn in KMnO4 is +7. Keep in Mind! S.Q. What is oxidation number of hydrogen and oxygen in different compounds. Ans. • Hydrogen shows -1 oxidation state with all metals. • Hydrogen shows +1 oxidation state with all non-metals. • Oxygen shows negative oxidation state with all metals. • Oxygen shows negative oxidation state with all non-metals except fluorine. • Oxygen shows -2, - 1 and -1/2 in H2O, Na2O2, and KO2, respectively. FINDING OXIDATION NUMBER OF AN ELEMENT IN A COMPOUND OR A RADICAL

Determining oxidation numbers in compounds and radicals General rule The oxidation number of any atom in a compound or radical can be determined by applying the standard rules. Compounds of a metal with a non-metal In such compounds, the metal always has a positive oxidation number and the non-metal has a negative oxidation number. Example: Sodium sulfide (NazS) Na = +1 (fixed for Group 1 metal 2(+1) + S= 0 +2 + S= 0 S = -2 Oxidation number of S = -2 (i) Compounds of a non-metal with a non-metal • The compound with two different non-metals sign of oxidation number depends on difference of electronegativity. • The more electronegative element is given a negative oxidation number. • The less electronegative element is assigned a positive value accordingly. Sample problem 10.1 Find the oxidation number of S in sulfur dioxide (SO2). 0=-2 (by rule) For 2 oxygen atoms: 2 x (-2) = 4 Equation: S+ 2(2 x-2) = 0 S= +4 Oxidation number of S = +4 (ii) Compound ions (Polyatomic ions) • Compound ions contain two or more different atoms. • The sum of oxidation numbers equals the charge on the ion. Example: Sulfate ion (SO4}) and the nitrate ion NO3!. Sample problem 10.2 Calculate the oxidation number of sulfur in SO4" 0=-2 Solution For 4 oxygen atoms: 4 × (-2) = -8 Equation: [Oxi No. of S] +4[Oxi No. of O]= -2 S+ (4 x-2) =-2 S= +6 S- 8= -2 The oxidation number of sulfur in SO4 is +6. Quick Check 10.110 (a) Define oxidation and reduction in terms of electron transfer with examples. Ans. Oxidation and reduction in terms of electron transfer: • Oxidation: Loss of electrons by an atom or ion. Example: Na → Nat + é (Sodium loses one electron - oxidized) • Reduction: Gain of electrons by an atom or ion. Example: Cl2 + 2e → 2CH (Chlorine gains electrons - reduced) (b) Define oxidation and reduction in terms of change in oxidation number with examples. Ans. Oxidation and reduction in terms of change in oxidation number: • Oxidation: Increase in oxidation number. Example: Fert → Fe3+ + e (Oxidation number increases from +2 to +3) • Reduction: Decrease in oxidation number. Example: Cu2+ + 2é → Cu (Oxidation number decreases from +2 to 0) (c) Determine Oxidation Number: Ans. (i) Oxygen (O) in Na2O Na = +1 (each), total +2 0 = x x+2(+1)=0→x=-2 Oxidation number of O= -2 (ii) Iodine (I) in ICk Cl = -1 (each), total - 3 1 = X x+3(-1)=0→x= +3 Oxidation number of I = +3 (iii) Nitrogen (N) in NOs 0 = -2 (each), total -6 N=x x+(-6)=-1→x= +5 Oxidation number of N= +5 (iv) Chromium (Cr) in KzCrz0, K = +1 (each), total +2 0 = -2 (each), total - 14 Cr = x (2 atoms) 2(+1) + 2x + (-14) = 0 . 2+ 2x-14 = 0→ 2x= 12→ x=+6 Oxidation number of Cr = +6 (v) Manganese (Mn) in KMnO4 K = +1 (each), total +1 O = -2 (each), total -8 Mn = x (1 atom) (+1) +x+(-8) = 0 x-7=0→x = +7 Oxidation number of Mn = +7 DISPROPORTIONATION REACTION

Disproportionation reaction Definition: A disproportionation reaction is a chemical reaction where a single substance acts as both the oxidizing and reducing agent, resulting in two different products with different oxidation states. Examples: Decomposition of Hydrogen Peroxide Reaction: 2H, O z(e) → 2H2O(e) + O2(g) Oxidation number of Compound oly gen - 1 H2O2 H2O - 2 (Reduction) 02 0 (Oxidation) Reaction of Chlorine with Sodium Hydroxide Reaction: 3Cl, + GNaOH -> 5NaCl + NaCeO, + 3H,0 Oxidation number of chlorine atom in Cl2 = 0 Oxidation number of chlorine in Clz = - 1 (Cl is reduced from 0 to - 1) Oxidation number of oxygen in H202 = - 1 Oxidation number of oxygen in H202 = - 3 (Cl is oxidized from 0 to +5) Oxidation number of oxygen in H2O = - 2 (oxygen is reduced oxygen is oxidized from d - 1 do 0) This is a classic disproportionation reaction of chlorine. OXIDIZING AGENT (OXIDANT) AND REDUCING AGENT (REDUCTANT) Oxidizing agent: An oxidizing agent (atom, ion or molecule) is that substance which oxidizes some other substance, and is itself reduced to a lower oxidation state by gaining one or more electrons. General form X (Ox. No. = 0) + e → X (Lower Ox. No.) Example: Cl + e → CI (oxidation number = - 1) lower oxidation number • Cl is reduced from 0 to -1 • Cl acts as an oxidizing agent General form Na → Nat + e • Na is oxidized from 0 to +1 • Na acts as a reducing agent Quick Check 10.2 (a) Identify the oxidized species and reduced species in the following equations. Also identify the oxidizing and reducing agents on the reactant side. Ans. Oxidized & reduced species, oxidizing & reducing agents (i) Reaction: Fe, 03(s) + CO (8) → FE(s) + CO2(g) Oxidized Species: CO → CO2 (Carbon is oxidized: C goes from +2 to +4) • Reduced Species: Fe3+ in FezO3 → Fe° (Iron is reduced from +3 to 0) • Reducing Agent: CO (it donates electrons and gets oxidized) Oxidizing Agent: FezO3 (it gains electrons and gets reduced) (il) Reaction: CH4(8) + 202(g) - Oxidized Species: C in CHa → C in COz (Carbon is oxidized: -4 to +4) Reduced Species: 02 → H2O (Oxygen is reduced: 0 to -2) • Reducing Agent: CH4 (it donates electrons) • Oxidizing Agent: 02 (it accepts electrons) (b) Identify the species in the following reaction which undergoing both oxidation and reduction. (i) Reaction: 2Na, 2(s) + 2H,0(8) - Oxidation Species Number of CI 0 CI2 Nacl - 1 (Reduction) +3 (Oxidation) NaCIO2 → CO2 (g) + 2H,0(.) →4NaOH (ag) + 02(g) O2 in Naz02: In NaUH, O is reduced (- 1 to - 2). In 02, 0 1s oxidized (- 1 to 0). (il) Reaction: → PН 3(в) + 3Н,PО 2 (aq) P4(3) + 30Н(ag) + 3H, (c) Phosphorus in PA: In PH3, P is reduced (0 to - 3) In H2P02 P. undergoes both is oxidation and reduction. Again, this is a disproportionation reaction. BALANCING OF REDOX EQUATIONS BY OXIDATION NUMBER METHOD

Carry out the following steps for balancing of redox equations by oxidation number method: (i) Skeleton equation: Write down the skeleton equation of the redox reaction under consideration. (ii) Change in oxidation number: Identify the elements, which undergo a change in their oxidation number during the reaction. (iii) Record: Record the oxidation number above the symbols of the element, which have undergone a change in the oxidation number (iv) Indicate oxidation number by arrows: Indicate the change in oxidation number by arrows joining the atoms on both sides of the equation. It shows number of electrons gained or lost. Equate the electrons: Equate the increase or decrease in the oxidation number, i.e., electrons gained or lost by multiplying with a suitable digit. (vi) Inspection method: Balance the rest of the equation by inspection (hit and trial) method Keep in Mind! Q. How change in oxidation number is predicted from chemical equation? (a) If an element in a species undergoes only increase in oxidation number or decrease in oxidation number in a reaction. The specie containing such element is written once on the RHS of the equation. ZM (s) + HCl (aq)ZnCl2(ag) + H2 (g) (b) If an element in a species undergoes both increase in oxidation number or decrease in oxidation number simultaneously in a reaction (as in case of self-redox or disproportionation reaction). The specie containing such element is written twice on the RHS of the equation First for increase in oxidation number while second is written for decrease in oxidation number. →5NaCl + 3H20 + NaClO 3 3Cl2 + 6NaOH (c) If an element in a species undergoes increase or decrease in oxidation number as well as no change in oxidation number in a reaction. The specie containing such element is also written twice. First, for change in oxidation number while second is for no change in oxidation number. HNO, + Cu + HNO, -Cu(NO3)2 + H2O+ NaOCl HCe+ K,CrO, + HCl→Cl, + KCl + CrCl, + H2O Sample problem 10.4 Balance the following equation by oxidation number method. K, CrO, + HCl→ Cl, + KCl + CrCl3 + H2O Solution (i) Write down the oxidation number of each element. (ii) Identify the elements that show change in oxidation number on both sides. be 2e Here, Cr undergoes a change in oxidation state from +6 to +3 and it is reduced acts as an oxidizing agent) while CrCl undergoes a change in oxidation state from -1 to and it is oxidized (acts as a reduction agent). (iii) Multiply Cl2 with 3 and Cr with 2 to balance the loss and gain of electrons. K, CrO, + HCl→3Cl, + KCl + 2CrC, + H2O (iv) Balance the remaining equation by inspection method (a) Balance K on both sides by multiplying KCl by 2. K_CIO, + HCl → →3Cl, + 2KCl + 2CrC, + H2O (b) To balance Cl atoms, multiply HCl with 14. K‚Cr,O, +14HCl-→3Cl≥+2KCl+2CrCl3 + H,° (c) To balance oxygen, multiply H20 by 7. K,CrO,+14HCl→3€l≥+2KCl+2CrCl3 +7H,° Did you know? S.Q. What is hit and trial method in balancing redox equations? Ans. Sequence of the balancing the chemical equation using inspection (hit, and trial) method. Firstly, balance atoms of all elements except nitrogen, oxygen and hydrogen. Secondly, balance atoms of nitrogen (if in the equation). Thirdly, balance atoms of oxygen (if in the equation). Lastly, balance atoms of hydrogen (if in the equation). Quick Check 10.3 Balance the given equations by oxidation number method. (i) Fe, 03(s) + CO(8) → FE(s) + CO2(g) (i) Mno, + HCe → MnCl, + H2O + Cl2 (iii) Fe* + MnO; → Fe** + Mn?+ (iv) K,CO, + FeSO, + H,SO, → C5 (SO4)3 + Fez (SO4)3 + K,SO, + H2O Ans. (i Fe, 0(s) + CO(8) → Fe(s) + CO2(g) Oxidation number changes: • Fe: +3 → 0 (reduction, gain of 3 electrons per Fe) • C in CO: +2 → +4 (oxidation, loss of 2 electrons) Change per formula unit: • 2 Fe atoms (from Fez03) gain 6 e- • 3 CO molecules lose 6 e- Balanced equation: Fe,O, + 3CO → 2Fe + 3CO, 1,O + Cl2 (il) MnO, + HCe → MnCl, + H Oxidation number changes: • Mn: +4 → +2 (reduction, gain of 2 e) • Cl in Hcl: -1 → 0 (oxidation, loss of 1 e per CI) To balance e: • 2CI lose 2 e → 1 Cl • 1 Mn gains 2 e Balanced equation: MnO, + 4HCl → MnCl, + 2H2O + Cl, (iii) Fe?* + MnO; → Fe** + Mn* (in acidic medium) Oxidation number changes: • Fe: +2 → +3 (oxidation, loss of 1 e) • Mn: +7 → +2 (reduction, gain of 5 er) To balance e ee. • 5 Fezt → 5 Fe3+ (lose 5 e) • 1 Mn04 → Mn2+ (gain 5 e) Also balance oxygen and hydrogen in acidic medium. Balanced equation (acidic medium): 5Fe* + MnO; + 8H* → 5Fe'* + Mn't + 4H,0 (iv) K CIO, + FeSO, + H,SO, → C (SO4)3 + Fe, (504)3 + K,50, + H2O Oxidation number changes: • Cr: +6 → +3 (reduction, gain of 3 e per Cr) • Fe: +2 → +3 (oxidation, loss of 1 e per Fe) From Cr202: 2 Cr gain 6 e- So, 6 Feżt needed (each loses 1 e) → 6 Fe3+ Balanced equation: K, CO, + 6FeSO, + 7H,SO, → C (SO4)3 + 3Fe2(504)3 + K,SO, + 7H2O ELECTROLYTIC CELL

Electrolytic Cell Definition: An electrolytic cell is a device that converts electrical energy into chemical energy through a process called electrolysis. Main components of an electrolytic cell (constructions). (i) Electrodes Made of metal or graphite (carbon). • Cathode (-): Negatively charged electrode. • Anode (+): Positively Anode charged electrode. (ii) Electrolyte: The electrolyte (Positive electrode) is the compound that is ionized, it is either. A molten ionic compound or solution containing aqueous free ions. • Carries electric current via 1on movement. (iii) Power Supply Provides direct current (DC) to drive the non-spontaneous reaction (iv) Structure of cell actually depends upon the nature of element extracted. Working: When direct current is applied, one electrode become positively charged (anode) and other become negative charge (cathode) Positive ions (cations) migrate to the cathode, where they gain electrons (reduction). Negative ions (anions) migrate to the anode, where they lose electrons (oxidation). • The movement of ions results in chemical changes at the electrodes and overall conversion of electrical energy to chemical energy. Process Electrode Charge Reduction Cathode Negative Positive Anode Oxidation Applications of electrolytic cells • Electrolysis of NaCl produces sodium metal and chlorine gas : Metal refining (copper purification) Electroplating of metals. • Production of caustic soda (NaOH) and other chemicals • Extraction of non-metals like chlorine • Purification of reactive metals REDOX REACTIONS IN ELECTROLYSIS Redox reactions in electrolysis The passage of electric current drive a non-spontaneous reaction in which compound is taken is called electrolysis. Electrolysis involves redox reactions, the oxidation and reduction. • Reduction occurs at the cathode (gain of electrons) • Oxidation occurs at the anode (loss of electrons) At the Cathode (-): Reduction happens Cations (positive ions) move to the cathode, they gain electrons (reduction) Products may be metal atoms or hydrogen gas, if hydrogen gas is formed it bubbles off. Examples: Direct current power supply → Cathode (Negative electrode) → Electrolyte Fig: Main parts of an electrolytic cell Ions move toward Reaction Positive ions (cations) Gain of electrons (e) Loss of electrons (e) Negative ions (anions) Copper ions in solution + 2e → CU(s) Copper metal is deposited at the cathode. Hydrogen ions from acidified water 2H(aa) + 2e → H2(g) Hydrogen gas bubbles off. At Anode (+): Oxidation happens Anions (negative ions) move to the anode. They lose electrons (oxidation) Products may be non-metal gases or other substances. Examples: Chloride ions 2Cl (aq) →Cl 2(g) + 2e Chlorine gas is released. Hydroxide ions (from water) 40H (ag) → 02(g) + 2H, 0(e) + 4è Oxygen gas is released at the anode Electrolysis of molten zine chloride (ZnClz): In the redox reaction of electrolysis of ZnCl2, the electrode reactions are Reaction Electrode Cathode (-) 15. Anode (+) 2Cl (e) Overall cell reaction ZnCl 2(0) → Zn(s) + Cl The electrons lost at the anode are equal to the electrons gained at the cathode, maintaining charge balance. Quick Check 10.4 Molten copper (II) bromide (CuBrz) is electrolyzed using inert electrodes. (i) Which ions are present in this solution? (il) Name the electrode on which the formation of copper metal occurs? (iii) Give the half-equation that represents the formation of copper metal. (iv) Write the half-equation for the formation of bromine on the other electrode. Ans. (i) Ions present in solution In molten CuBiz, it dissociates into: Cuz* and Br ions (only these ions, as it is molten (ii) Electrode on which the formation of copper metal occurs Cathode: Copper metal is formed at the cathode because Cu2+ gains electrons in reduction. (iii) Half-equation for the formation of copper metal: Cu2+ + 2e → Cu (Reduction at cathode) (iv) Half-equation for the formation of bromine: 2Br → Br, + 2e (Oxidation at anode) Type of reaction Reduction = →Cl Oxidation (g) + 2e -not aqueous MASS OF A SUBSTANCE DEPOSITED DURING ELECTROLYSIS

Mass of a substance deposited during electrolysis The mass of a substance deposited (or liberated) during electrolysis is directly proportional to: • The time over which constant current flows, and the strength of electric current. • The number of electrons involved in the electrode reaction. Combining current and time. The total charge passed is calculated using: 0=1xt Where: • Q= Charge in coulombs (C) • I= Current in amperes (A) • t = Time in seconds (s) The mass of substance produced at (or removed from) an electrode during electrolysis is • 1 Faraday (F) = 96,500 C/mol Electrolysis of Silver Nitrate (AgNO), sliver is deposited at cathode. A8(a4) + e → AB (s) 1 mole of electrons (1 Faraday) deposits 1 mole of silver (107.9g). • Charge required = 1 × 96,500 C Electrolysis of Copper(II) Sulfate (CuSO4), copper is deposited at the cathode. Cu (24) + 2é → CU(s) Imol moles 2 moles of electrons (2 Faradays) are needed to deposit 1 mole of copper (63.5g). • Charge required = 2 × 96,500 C = 193,000 C General Formula (Using Moles and Faraday) No. of Faradays=No. of moles of electrons (e) gained or lost. AMOUNT OF SUBSTANCE PRODUCED DURING ELECTROLYSIS The value of F can be used to calculate the mass of substance deposited at an electrode and the volume of gas produced at an electrode. Sample problems 10.5 Calculate the mass of lead (P) deposited at the cathode during electrolysis when a current of 1.50 A flows through molten lead (II) bromide (PbBr2) for 20.0 min. (relative atomic mass, Ar value: [Pb] = 207; F = 96500 C mol") Solution: Step 1: Write the half-equation for the reaction. Pb+2 (29) + 2é Pb (s) Step 2: Find the number of coulombs required to deposit 1 mole of product at the electrode. 2 moles of electrons are required per mole of Pb formed 1mol = 2F = 2 × 96500 = 193000 C mol-1 Step 3: Calculate the charge transferred during the electrolysis. Step 4: Calculate the mass by simple proportion using the relative atomic mass. 193000 C deposits 1 mole Pb, which is 207 g Pb 207 -×1800 = 1.93g of Pb So, 1800 C deposits = - 193000 AVOGADRO'S CONSTANT BY THE ELECTROLYTIC METHOD

Avogadro constant and electrolysis The Avogadro constant Na is the number of particles (atoms, ions, or molecules) in 1 mole of a substance. The electrolytic method allows us to determine L by calculating the charge associated with 1 mole of electrons and dividing it by the charge on a single electron. Charge on 1 mole of electrons NA =- Charge on 1 electron The charge of an electron can be calculated experimentally. • Charge on 1 mole of electrons = 1 Faraday = F = 96,500 C/mol • The results show the charge on electron is approximately 1.60×10-19 C FINDING THE CHARGE ON 1 MOLE OF ELECTRONS Objective To calculate the charge carried by 1 mole of electrons using a simple electrolytic experiment involving copper(II) sulfate solution and copper electrodes. Apparatus Required • Pure copper cathode and copper anode • Aqueous copper(Il) sulfate (CuSO4) • Power supply • Variable resistor (to control current) • Ammeter and stopwatch Ammeter Distilled water and propanone for cleaning electrodes) • Electronic balance Copper anode < Procedure (i) Weigh the copper cathode and anode before electrolysis. (ii) Set up the electrolytic cell with CuSO4 solution. (iii) Use a variable resistor to maintain a constant current (0.2 A). (iv) Pass the current for a fixed time (40 minutes = 2400 seconds). (v) After electrolysis: Remove the electrodes, wash with distilled water, then with propanone and dry and reweigh both electrodes. = |x t = 1.50 × 20 × 60 = 1800 C Battery Variable resister → Copper cathode → CuSO Arag) solution Fig: Apparatus for calculating the mass of copper deposited Observations • The cathode gains mass (copper is deposited). • The anode loses mass (copper dissolves into solution as Cu2+). • Prefer using the mass loss of the anode for accurate results as copper may not stick perfectly to the cathode). Calculating charge on an electron A sample calculation is shown below, using a current of 0.20 A for 34 min. Mass of anode at start of the experiment = 56.530 g Mass of anode at end of experiment = 56.394 g Mass of copper removed from anode = 0.136 g = 1xt Quantity of charge passed Q = 0.20 × 34 × 60 = 408 C As the amount of copper removed is 0.13 g = 408 C So, 0.13 g of copper requires 408 1 g of copper requires 0.136 408 = - -x 63.5 63.5 g of copper requires 0.136 =190500 C But the equation for the electrolysis shows that 2 moles of electrons are needed to produce 1 5JP mole of copper: Cư(aq) + 2e →CU(s) The charge on 1 mole of electrons = - = 95,2500 2 If the charge on one electron is 1:60 x 10-19 C, 95250 Charge on an electron = - 6.022x1025=1.58x10-19C This is in good agreement with the accurate value of 6.6.02 × 10^23 mol-1 Hence Avogadro's number is calculated by electrolytic method. Quick Check 10.5 (a) An aqueous solution of silver nitrate is electrolyzed. Calculate the mass of silver deposited at the cathode when the electrolysis is carried out for exactly 35 main using a current of 0.18 A. (Ar [Ag) = 108; F = 96500 C mol). Ans. Electrolysis of silver nitrate (AgNO3) We are given • Time = 35 min = 35 × 60 = 2100 seconds • Current (I) = 0.18 A • Ar of Ag = 108 • Faraday's constant (F) = 96,500 C/mol • Silver ion is Ag*, so 1 mole of Ag+ gains 1 mole of electrons (n = 1) Step 1: Calculate charge (Q) passed 2= 1xt= 0:18x2100=378 C Step 2: Use formula to calculate mass of silver deposited mass = ex Ar 378×108 mass = 1x 96500 Answer: 0.423 g of silver is deposited at the cathode. (b) An electric current of 1.04 A was passed through a solution of dilute sulfuric acid for 6.00 mln. The volume of hydrogen produced at STP was 41.5 cm?. (i) How many coulombs of charge were passed during the experiment? (i) How many coulombs of charge are required to liberate 1 mole of hydrogen gas? (F = 96500 C mol) Ans. (i) Electrolysis of dilute sulfuric acid Given: • Current (I) = 1.04 A • Time = 6.00 min = 6 × 60 = 360 s 2= 1xt=1.04x360= 374.4 C Ans. 374.4 C of charge passed (ii) In electrolysis of H2O: 2H* + 2é → H2 To produce 1 mole of H2, 2 moles of electrons are required Charge required = 2 x F=2 x 96,500 = 193,000 C Ans. 193,000 C are required to liberate 1 mole of H2 gas ELECTRODE POTENTIALS

Electrode potential Definition: When a metal is dipped into solution of its own ions, the electrical potential is established between metal and metal ions in the solution. It is electrode potential and it indicates the case of oxidation or reduction of substance. a dynamic equilibrium exists between two chemically related During redox processes, species in different oxidation states. Example: When a copper rod is dipped into solution of its ions. CU(29) + 2e = CU(s) This equilibrium involves two opposing half-reactions: Opposing reactions at the electrode (a) Oxidation (Loss of electrons): Parent atoms converting into metal ions from the electrode. • Copper metal (Cu) loses electrons to form Cu2+ ions: • Electrons remain on the surface of the electrode. (b) Reduction (gain of electrons): Ions acting the electrons from metal electrode and get, deposited as metal atom on cathode surface • Cu2+ ions in solution gain electrons to form solid copper: Cu2+ 1(aq) + 2e - n× F 40824 • = - = 0.423 g 96500 (aq) + 2e Redox equilibrium is established When: • The rate of oxidation becomes equal to rate of reduction. • This balance defines the electrode potential of the metal. • A voltage (potential difference) is established between: The metal electrode and the solution of its ions. • This is called the electrode potential, and it reflects the ease with which a substance can be oxidized or reduced Comparison of metals (i) Unreactive metals (Copper) • The redox equilibrium lies to the right: Cu2+ (ag) + 2E →CU(s) Cu2+ ions are easily reduced, they readily accept electrons, if equilibrium is compared with equilibrium setup by other metals. • Copper has a higher electrode potential. (i) Reactive metals • The redox equilibrium lies to the left: Vanadium further V2+ (2q) + 2é - • Vt ions are harder to reduce, i.e., they accept electrons less readily. • Vanadium has a lower electrode potential. ELECTRICAL DOUBLE LAYER Absolute electrode potentials • The absolute electrode potential cannot be measured because of formation of electrical double layer when an element is dipped in a solution of its own ion. • When a metal is placed in a solution containing its own ions + 2e • Zinc atoms lose electrons, forming Zn2+ ions that enter the solution. • This process leaves excess electrons on the surtace of the metal. Formation of the Electrical Double Layer • Zn2+ ions accumulate in the solution near the metal surtace, they are attracted back to the surface of metal. • This creates charge an electrical potential between metal and metal ions in the solution. • This is called the electrical double layer. Figure shows the formation of double layer forms between the metal and solution interface. • The potential cannot be measured directly due to formation of double layer. • The difference in potential of metal ion system and another system can be measured. • It is called electrode potential and measured in volts. • So comparison system is used to measure electrode potential called SHE standard hydrogen electrode. (s) • Zinc rod +2 +2 +2 ZMSO 4(a9) +2 Zinc(II) ions Fig: Formation of electrical double layer Concept Forms due to separation of charges at metal-solution Electrical double layer interface. Internal charge buildup prevents isolation of a single Absolute potential can't electrode's voltage. be measured STANDARD HYDROGEN ELECTRODE

Standard hydrogen electrode (SHE) The SHE is a reference half-cell used to measure the standard electrode potential (E°) of other electrodes. It is assigned a value of 0.00 V by definition. Components of the SHE (i) Hydrogeti Gas (H2) at 1 atm (101 kPa) pressure. (i) Hydrochloric acid or any solution providing Ht ions at 1.00 mol dm (ili) A platinum electrode coated with platinum black, contact with H2 gas and Ht ions. • A finally divided platinum provide a surface for electron exchange to H2 gas and Ht! ions in solution. • It speedup the establishment of free if" equilibrium. • It is inert electrode. • Standard electrode potential are all half-cells (E°) for measured in comparison to this Glass 'bell' with electrode. a hole in for H, to • SHE is connected to another bubble out on half-cell, the reading voltmeter gives the standard electrode potential for that half- cell. • The Half-equation for the hydrogen electrode can be written. 2H + +2e = It explains the hydrogen gas and hydrogen ions have equal to lose and gain electron. Feature Gas used Electrolyte Electrode Potential Function Explanation - Hydrogen gas supply at l atm pressure → Platinum wire → Ht (1mol/dmt3) → Platinum electrode Fig: Standard hydrogen electrode (SHE) 2(g) Description H2 at 1 atm 1 M Ht (e.g., Hcl) Platinum coated with Pt black E° = 0.00 V (by definition) Reference to measure E° of other cells STANDARD ELECTRODE POTENTIAL

Standard Electrode Potential (E°) Definition: The standard electrode Potential (E°) is the voltage of a half-cell, measured under standard conditions, with a standard hydrogen electrode (SHE), as the other half-cell. It reflects a substance's tendency to gain or lose electrons, to be reduced or oxidized. Standard Conditions • The voltage of electrochemical cell depends on concentration, temperature and pressure of the gas. • To ensure fair comparison, the standard conditions these must be strictly maintained, to measure electrode potential • Under these conditions, the electrode potential measured is called standard electrode potential. Its symbol is E° and is spoken as 'E standard'. Condition Ion concentration Temperature Gas pressure Reference electrode Symbol used Purpose of E° • Compare the oxidizing/reducing power of different electrodes. • A more positive E° means a stronger tendency to gain electrons (reduction). • A more negative E° means a stronger tendency to lose electrons (oxidation). MEASURING STANDARD ELECTRODE POTENTIALS Measurement of standard electrode potential (E°) There are three main types of half-cell whose E° value can be obtained when connected to a standard hydrogen electrode (SHE) Types of half-cells (i) Metal / Metal ion half-cell • A metal electrode (e.g. copper strip) is placed in a solution of its ions (e.g. Cu2+). • Connected to SHE, the measured voltage is +0.34 V • This means copper has a greater tendency to gain electrons (be reduced) than hydrogen. Example: Cu2+/Cu(s) Half-equation: → CU (s) Cula + 2e - E° = +0.34 V (ii) Non-metal / non-metal ion half-cell • Chlorine gas is bubbled over a platinum electrode in a solution containing Clions. • The platinum electrode is inért and conducts electrons. Example: Clz(g) Cl (aq) blue 1.00 mol dm ? 25°C (298 K) 1 atm (101 kPa) Standard Hydrogen Electrode (SHE) E° Half-equation: Cl 2(g) + 2é → 2Cl (ag) (iii) Ion / ion half-cell Example: Fe3t/Fe2t • This uses a platinum electrode in a solution containing both Fert and Fe3+ ions. • No metal is involved; the platinum acts as a conductor. E° = +1.36 V Half-equation: Fe(ag) te → Fe (ag) E° = +0.77 V Measurement of E° of Cu(s) Cu(a) • Connect the half-cell of Cu2+ / cu to the SHE using a salt bridge to allow ion flow. • Use a voltmeter to measure the cell potential. • If the SHE is the negative terminal and the other half-cell is positive, the reading is the • For Cu2+/Cu, the measured potential is +0.34 V, meaning copper ions readily gain electrons (are easily reduced). Copper Half-Cell (Cu2+/Cu) voltmeter ple 519 + 0.34 V e Salt bridge Copper rod < Cu 298 K Cut2 Temperature (1 mol/dm?) Fig: Measuring the standard electrode potential of Cu2+/Cu half-cell Reaction at copper half-cell: + 2é CU(s) aq) E° = +0.34 V Reaction at hydrogen half-cell (SHE): ZH (ag) + 2e F = H2(g) E° = 0.00 V Conclusion: Cu2+ ions are more likely to gain electrons than H*. ions. H2 (g) 1 atm H2 (1 mol/dm3) → Platinum Cu* accepts electrons from H*/H, half-cell and H2 will lose electron to the Cu2+/Cu half-cell. Measurement of E° of Zn*/Zn Zine Half-Cell (Zn/Zn) voltmeter e+ 0.76 V Salt bridge Zinc rod < Zn 298 K Znt2 Temperature (1 mol/dm) Fig: Measuring the standard electrode potential of a Zn?* /Zn half-cell The voltage of the (Zn2+/Zn) half-cell is -0.76 V Reaction at zinc half-cell: + 2é = E° = - 0.76 V Reaction at hydrogen half-cell (SHE) : H2(g) 2H (ag) + 2e = E° = 0.00 V Conclusion • Zn2+ has a lower (more negative) E°, so it's harder to reduce than Ht. • Zinc metal is oxidized, giving up electrons to the Ht/H2 half-cell. • Zinc is the negative terminal, and Ht is reduced to H2. • Ht ions gains electrons from the Znt2/Zn half-cell. ELECTROCHEMICAL CELL (GALVANIC CELL) 012. Describe the construction and working principle of the electro chemical cell (Galvanic cell). Ans. Galvanic (Voltaic) Cell A cell in which the conversion of chemical energy into electrical energy takes place spontaneously in exothermic redox process with oxidation at anode and reduction at cathode. Cell representation: A galvanic cell is designed to take advantage of spontaneous transfer of electrons. Cu2+ / Cu || Zn? / Zn cell E° = 1.10 V + e- H2 (g) 1 atm →H* (1 mol/dm?) → Platinum . Construction: Two half-cells are connected together, to have a electrochemical cell (galvanic cell). Voltage of this cell. A Cu2+/Cu half-cell connected to a Zn?+ /Zn half-cell to make a complete galvanic cell in which voltage is measured. Half-cells are connected together using: • Wires connecting the metal rods in each half-cell to a high-resistance voltmeter: round the electrons flow this external circuit from Copper rod the metal with the more negative (or less positive) electrode potential to the metal with the less negative (or more positive) electrode potential. • A salt bridge to complete electrical (1 mol/dm') allowing the movement of ions between the two half cells so that ionic balance is Fig: An electrochemical cell, made by connecting a maintained a salt bridge Cu2+/Cu half-cell to a Zn?+ /Zn half-cell movement of electrons. It prevents accumulation of charges. • A salt bridge can be made from a strip of filter paper (or other inert porous material) soaked in a saturated solution of potassium nitrate. Voltage of Cell: The voltages for the half cells can be represented by the following halt- equation: =CU(s) Cu(ag) + 2é = Zn(a9) + 2e = Z (s) • Working an electrochemical cell is made by connecting a Cu2+/Cu half-cell to a Zn2+/Zn half-cell The relative values of these voltages tell us that Zn2+ ions are more difficult to reduce than Cu2+ ions. At Anode: Zinc undergoes oxidation, losing electrons to form zinc ions. 2M (s) At Cathode: The electrons flows from anode to cathode. Copper ions gain electrons and are reduced to form copper metal. Cu(a) + 2é - Overall Reaction: The overall cell reaction can be represented as: ZM (s) + Cu(ag) →→ Zma The process effectively converts the chemical energy of the redox reaction into electrical energy, which can be connected to perform work. The voltage generated by this cell is +1.10 V. Equation Ecell = Epe oxi = + 0.34-(-0.76) = +1.10 red. + E° In a Cu-Zn galvanic cell, zinc (Zn) serves as the anode, and copper ions (Cu2+) in solution serve as the cathode. High resistance voltmeter - e- e-+ Salt bridge Zinc rod (1 mol/dm ) E° = +0.34 V E° = -0.76 V →CU (s) (aq) + CU(s) Other Examples The E° value for the Ag /Ag half-cell is more positive than for the Zn?* /Zn half-cell. So, the Ag*/Ag half-cell is the positive pole and the Zn?+/Zn half-cell is the negative pole of the cell. Part Anode Zinc rod (oxidation occurs here, negative terminal) Cathode Copper rod (reduction occurs here, positive terminal) Filter paper soaked in KNO - allows ion flow to balance Salt Bridge charges but not electron flow Electrons flow from Zn to Cu through a wire, powering a External Circuit device (e.g., bulb, motor) Measures the cell potential (1.10 V in this case) Voltmeter Quick Check 10.6 (a) What is the function of a salt bridge in the voltaic cell? Ans. Function of salt bridge: • It completes the circuit by allowing the flow of ions. • It maintains electrical neutrality by transferring ions between the two half-cells. • It prevents the mixing of different solutions that could cause a direct reaction. The salt bridge maintains charge balance and allows the cell to function continuously. (b) Calculate the voltage of a cell with iron and copper electrodes: Given E° = -0.44 V • Fet + 2e → Fe(s) E° = +0.34 V • Cu2+ + 2e → Cu(s) Ans. In a voltaic cell • The metal with the higher Ered Value acts as the cathode (reduction). • The metal with lower E red value acts as the anode (oxidation). • Anode (oxidation): Fe → Feźt + 2é- • Cathode (reduction): Cu2+ + 2e → Cu E° = + 0.34 V E° cell = Ecathode tEa, Cell voltage (Ecell): 'cell (0.34) + (0:44)=0.78V Answer: The voltage of the cell is 0.78 V APPLICATIONS OF E" VALUES

Applications of E° values (i) Direction of electron flow Direction of flow in wires of external circuits can be deduced by comparing E° values two half cells that make electrochemical series. Example: Ag (a9) +e = Description E° = + 0.44 V anode E° = + 0.80V E° = - 0.76V Electrons flow from: • Zinc (Zn) electrode because it has the more negative E° value (-0.76 V), indicating it loses electrons more readily (oxidation). Electrons flow to: • Silver (Ag+) electrode because it has the more positive E° value (+0.80 V), indicating it gains electrons more readily (reduction). Direction of electron flow (External Circuit): Electron flows from Znt2/Zn (anode) to Ag+/Ag (cathode) or from negative pole to positive pole. Fundamental rule in electrochemistry: Electrons always flow from the anode to the cathode, from the more negative E° to the more positive E° (ii) Feasibility of a reaction using E° values Standard electrode potential E° gives us the measure of how easy or difficult it is oxidized or reduced. E° values are written as reduction half-equations by convention. For each half equation, more oxidized form is on the right side and more reduced form on the right. To check the feasibility of reaction between • The more positive E° value, the greater tendency for the half equation to proceed in forward direction, easier it is to reduce the specie on the left of half equation. The more positive the E° value, the stronger the oxidizing agent. • The less positive E° values, the greater tendency for the half equation to proceed in reverse direction, easier it is to oxidize the specie on the right of the half equation. The less positive (or more negative) the E° value, the stronger the reducing agent. Reaction Example: Mg?+ + Ag → ? We need half-cell equations and E° values E° = -2.32V Mg?*+2e ==Mg E°= +0.80V Ag*+e ==Ag So, E° cell = E°red - E°oxi E°cell = (-2.32) - (0.80) =-1.52 V → Not feasible (reaction not spontaneous) Reaction Example: Ag+ + Mg →? Negative value of E° cell indicates the reaction is not feasible. On the other hand reaction between Ag and Mg is feasible as for this reaction E°cell will be positive as given below. E°cell = E red - E° oxi = 0.80 - (-232) E°cell = + 1.52 V The feasible reaction is: 2А 8(2) + Mg (s) → Mg(24) + Ag(s) Strongest Oxidizing agent 2Claq) E° = +1.36 V Ch+ 2é = 18(s) A8 (ag) tè Cuta) + 2e == Cus) H (ag) tè =2#2181 Zn(aa) + Zé (Tendency to gain increasing for ions/molecules on left of electrons is increased = Mg (s) Mg (ag) + 2è Fig: Standard electrode potentials for some oxidizing and reducing agents Quick Check 10.7 (a) Explain the following keeping in view the standard electrode potential values (E° Clz/2Cl' = +1.36 V E° (Brz/2Br) = +0.54 V) (i) The aqueous solution of iodide ion can be oxidized with bromide or chloride ions. (ii) Why bromine does not react with chloride ions? Ans. Given standard electrode potentials (E°): • E°(Cl/2CI) = +1.36 V • Е°(Brz/2Br) = +1.07 V • E°(1z/2I) = +0.54 V Higher E° = Stronger oxidizing agent (greater tendency to gain electrons) Lower E° = Stronger reducing agent (greater tendency to lose electrons) Explanation: • iodide ion (IF) can be oxidized to l (I → 12), in the presence of chloride or bromide because they stronger oxidizing against than iodide. • Bromine (Brz) has E° = +1.07 V, which is greater than I2 (E° = +0.54 V). • So Brz can oxidize I to Iz. Because Brz is a stronger oxidizing agent than 12. (ii) Why bromine does not react with chloride ions? Ans. Explanation: Chloride ion (CI) has E° = +1.36 V for Clz/2C1 Bromine (Brz) has E° = +1.07 V For a redox reaction to occur The oxidizing agent must have a higher E° than the reducing agent. But here: Brz < Clz in E° → So Brz cannot oxidize CI to Cl Bromine does not react with chloride ions because it is a weaker oxidizing agent than chlorine. E° = +0.80 V E° = +0.34V E° = +0.00V electrons is increased E° = +Q.76V ions/molecules on right E° = -2.38V Stroegest Reducing agent (ill) Oxidizing agents and reducing agents using E° Values • Oxidizing agents are substances that gain electrons (they get reduced). • The more positive the E° value of a half-equation, the stronger the oxidizing agent. • So, species on the left side of half-equations with high positive E° values are strong oxidizers. • In reaction of cu and Znt2. Cu will not reduce Znt2 to Zn. Because Znt2 is stronger agent than cu. • A stronger reducing agent which should have higher negative E° value than E° value of Znt2/Zn can reduce the Znt2. So Mg is suitable reducing agent for Zn2+? E°=-0.76 V + E → ZM(s) Mg(ag) + 2e → Mg(s) E°=-2.38 V Reducing agents and E° values • Reducing agents are substances that lose electrons (i.e., they get oxidized). • The more negative the E° value of a half-equation, the stronger the reducing agent. • Species on the right side of half-equations with very negative E° values are strong reducers. Example E° = -2.38V M8(4) +26 → Mg(s) Mg is a very strong reducing agent much stronger than Zn. Zn is better oxidizing agent. • Oxidizing agents are substances that gain electrons and get reduced in the process. • These substances have higher (more positive) standard electrode potentials (E° values). Example of compounds as oxidizing agents • Potassium permanganate (KMnO4) acts as a moderate oxidizing agent. E° = + 0.54 V • Potassium dichromate (K,Cr,O,) acts as a very strong oxidizing agent due to its high E° value. E° = +1.33 V Example of compounds as reducing agents • Reducing agents are substances that lose electrons and get oxidized. • These have lower (more negative) E° values. • Potassium iodide (KI) • Iron(II) sulfate (FeSO4) VARIATION OF E° VALUE WITH ION CONCENTRATION

Variation of E° value with ion concentration • The position of an equilibrium reaction is affected by changes in concentration, temperature and pressure. • If we change the concentration or temperature of half-cell X, the electrode potential also changes. Standard electrode potential (E°) is measured under standard conditions: • Ion concentration = 1.00 mol dm3 • Temperature = 25°C (298 K) • Pressure = 1 atm When the ion concentration is changed, we no longer use E°, but rather E (non-standard electrode potential). Effect of ion concentration on electrode potential The position of redox equilibrium shifts when ion concentration changes, which affects the electrode potential. This can be explained using Le-Chatelier's Principle. • Increase in concentration of species on the left side of the half-equation → equilibrium shifts right → E becomes more positive / less negative. • Decrease in concentration of species on the left side → equilibrium shifts left → E becomes more negative / less positive. Zn2+/Zn Half-Cell Z(ag) + 2e → Zn(s) Standard E° = - 0.76 V Table 10.1 Variation of Potential with Ion Concentration [Fe?| (mol dm3) (Pet (mol dn") 1.00 > 1.00 1.00 < 1.00 1.00 > 1.00 1.00 < 1.00 → ZM (s) Zm (24) +e Fe3+/Fet Half-Cell Felt + e 1.00 mol dm 1.00 mol dm NERNST EQUATION 015. Write detail note on NERST equation to explain variation of electron potential. Ans. Nernst equation The effect of concentration and temperature on the value of E° Nernst equation. Consider a cell made up from a silver/silver ion electrode and a copper/copper(II) ion electrode, the reaction taking place is: CU(s) + 2A8 (a) → Cu(a) + 2Ag(s) Effect E° (V) + 0.85 More positive + 0.70 Less positive + 0.70 Less positive + 0.85 More positive E° = -0.76V Fe?+ E° = +0.77V cell can be deduced using the E° = + 0.46V When E° •Cell is plotted against concentration a graph is obtain as shown in figure. 0.5- 0.4 - ≥ 0.3- 0.2- 0.1 0 - 5 -8 -6 -7 log [Ag ] Fig: Increasing the concentration of silver ions in the cell reaction 'Cell (non-standard conditions for the cell as a whole) against A graph is plotted the value of E the logarithm of the silver ion concentration. The above redox cell reaction makes the value of cell more positive. For a given electrode Cu(s) tree Im The relationship is: E = E° + The Nernst equation relates the electrode potential (E) of a half-cell or full electrochemical cell under non-standard conditions to: • The standard electrode potential (E°) • The temperature • The concentration of ions involved in the reaction Where, E is the electrode potential under non-standard conditions. E° is the standard electrode potential R is the gas constant, 8.314 JK mol T is the kelvın temperature z is the number of electrons transferred in the reaction F is the value of the Faraday constant in C mol In is the natural logarithm oxidized reters to the concentration of the oxidized form in the halt-equation [reduced refers to the concentration of the reduced form in the half-equation. For a metal/metal ion electrode, we can simplify this equation in three ways: The natural logarithm, ln, is related to log to the base 10 by the relationship ln x = 2.303 log10 x At standard temperature the values of R, T and Fare constant The equation then becomes 0.059 E=E°+ Z «EO 0.46 V cell - 1 -0 - 4 - 3 - 2 [Reduced from] [reducedform] Sample Problem 10.6 What is the electrode potential of a Zn electrode dipped in a solution containing Zn? ions (0.1M), the e° value of Zn't / ZnO is -0.76V Using Half equation and E°, → Zn (a) Zn(ad) + 2é - E = (-0.76) + E = -0.76 + (-0.029) = -0.78V Since, concentration of Zn?+ is decreased to shift equilibrium backward and electrode potential decreases to -0.78V ACTIVITY SERIES OF METALS

Activity series The activity series of metals is a ranking of metals based on their reactivity, particularly their tendency to lose electrons and undergo oxidation. (i) Highly reactive metals • Metals higher in the activity series have more negative standard reduction potentials. • They are more likely to lose electrons and undergo oxidation. • Such metals act as reducing agents. • Metals at the top of the series are the most reactive with water, oxygen and acid. Example: Alkali metals like cesium (Cs) and sodium (Na) are highly reactive and oxidize readily, reacting vigorously with substances like water of to form ions and release hydrogen gas. Reaction of sodium with water 2Na + 2H,0 → 2NaOH + H, T (ii) Least reactive metals • Metals lower in the activity series, such as noble metals like gold and silver, do not oxidize easily. • Metals at the bottom of the series are the least reactive with water, oxygen and acid. Example: Metals such as gold, platinum and silver are least active metals because they are lowest in the series from metals. Reactivity series • The position of a metal in the activity series directly correlates with its ease of oxidation and overall reactivity. • Activity series of metals is actually reactivity series. • The activity series can be understood in terms of standard reduction potentials. • Metals with more negative potentials are more reactive and oxidize more readily. • Metals with more positive potentials are less reactive and oxidize less readily as given in Table10.2. E° = - 0.76V 8.31 × 298 - ln(0.1) 2x96500 Conclusion The activity series helps predict: • Metal displacement reactions • Reactivity with water and acids • Whether a redox reaction will occur It's a practical tool in electrochemistry, metallurgy, and industrial extraction of metals. High in Series (Top) More reactive More easily oxidized Have more negative E° Strong reducing agents React vigorously with water, acids Table 10.2 Activity Series of Metals Ions Metals E" (V) Li Lit - 3.05 - 2.93 KI+ Ba + Ba - 2.90 Srt Sr - 2.89 - 2.87 Cat Na t 2.71 freelm - 2.37 Mg?t Al3+ - 1.66 0.76 Zn?+ Crt - 0.74 Fet - 0.44 Cd+ - 0.40 Co2t - 0.28 Ni2+ - 0.25 Sn?+ - 0.14 Pb2+ Pb - 0.13 H2 H* 0.00 Cu Cu2+ +0.34 Hg +0.92 Hg?+ +0.80 Ag!+ Ag P+2+ Pt +1.18 Au'+ Au +1.50 Low in Series (Bottom) Less reactive Less easily oxidized Have more positive E° Weak reducing agents / act as oxidizing agents Do not react easily Trend Reaction Occurring React with cold water, replacing hydrogen 2M(s) + 2H20(1) - 2MOH(aq) +H2(g) React with steam, but not cold water, replacing hydrogen. Ms) +H20(i) - MO(ag) + H2(g) Do not react with water. React with acids, replacing hydrogen. Ms)+Hcl (aq) MC(ag) +H2(g) Included as reference Unreactive with water or acids. FEASIBILITY OF REDOX REACTIONS FROM ACTIVITY SERIES OR REACTION DATA

Feasibility of redox reactions from activity series data The activity series ranks metals by their reactivity, from most reactive (top) to least reactive (bottom). A metal higher in the series can displace a metal lower in the series from its salt solution. This allows us to predict if a redox reaction is feasible. Rule for feasibility: • A more reactive metal (higher in the series) can reduce the ion of a less reactive metal. • If the overall cell potential (Ecall) is positive, the reaction is feasible. • If Ecel is negative, the reaction is not feasible. Step-by-step process to check feasibility: (i) Write half-reactions for oxidation and reduction. (i) Look up standard electrode potentials (E° values). (ili)Calculate E°cell: Interpret the result: • E° 'Cell > 0 then reaction is feasible EcellO then reaction is Half-Reactions: Conside compete and from activity series, i m Fe** +2e =Fe(s) E(ox) =- 0.44 V Ag* + le ==Ag(s) Elox =+ 0.80V Overall reaction: → Fe(ag) + 2Ag(s) Fe(s) + 2A g(ag) Cell potential E° cell = 0.80 - (-0.44) = +1.24% The cell potential is positive so the reaction is feasible. Consider another example from the activity series of metals Al+ and Ag Al* + 36 = Al (s) E(red) = - 1.66V Ag*+ e==Ag(s) Elox= + 0.80V The overall reaction is Al* (aq) + Ag(s) → Al(s) + A8(a9) Cell potential Ec cell =-1.66 + 0.80 = - 0.86V The cell potential is negative so the reaction is not feasible. Conclusion: • Use the activity series or E° values to predict feasibility. • If a metal is higher in the series, it can replace a metal lower in the series. • Positive E°cell confirms the reaction can proceed spontaneously. Quick Check 10.8 (a) State the Nernst Equation, why is it significant in electrochemistry. Ans. Nernst Equation: For a half-cell reaction: 0.0591 [Red.form] E= E°-. - log n [Oxi.form Or for a general redox reaction: 0.0591 E = E° -log2 n Where: • E = electrode potential at non-standard conditions • E° = standard electrode potential • n = number of electrons transferred • [Mnt] = concentration of ions • Q = reaction quotient Significance: • It allows us to calculate electrode potential under non-standard conditions. • Shows how concentration of ions affects cell potential. (b) What is the effect of variation in ion concentration on the standard electrode potential of a half reaction? Ans. Effect of Ion Concentration on Electrode Potential • If ion concentration increases, the electrode potential increases for a reduction reaction. • If ion concentration decreases, the electrode potential decreases. Conclusion: Electrode potential depends on the concentration of ions in solution, as shown by the Nernst equation. (c) Give answers using E° values (EAg+/Ag = + 0.80 V, E°Crz+Cr= - 0.91 V, E®Fez+/Fe = - 0.44 V.) (i) Which of Ag+, Crit and Fett, is the strongest oxidizing agent? Which is the least oxidizing? Explain with reason. (ii) Arrange Ag, Cr and Fe in increasing order of their reducing powers. Ans. Based on E° values: • E°(Ag/Ag) = +0.80 V • E°(Fet/Fe) = -0.44 V • E°(Cr2+/Cr) = -0.91 V (i) Stronger oxidizing agent = higher E° • Ag+ has the highest E° → strongest oxidizing agent • Crit has the lowest E° → weakest oxidizing agent Answer: • Strongest oxidizing agent: Ag+ • Least oxidizing agent: Cr2t (ii) Arrange Ag, Cr, Fe in increasing order of reducing power • Reducing power increases with more negative E° Order of E° values: • Ag: +0.80 V • Fe: -0.44 V • Cr: -0.91 V Answer: Ag < Fe < Cr (increasing reducing power) (a) What is the electrode potential of a Cu electrode dipped in a solution containing Cu2+ ions (0.01 M), The E° value of Cu2+/Cu/ is + 0.34 V. Ans. Electrode potential of Cu in 0.01 M Cu2+ solution Given: • E°(Cu2+/Cu) = 0.34 V • [Cu+]= 0.01 M • n = 2 (2 electrons involved) Use the Nernst equation: 1 0.0591 E=E° - - 10g n [Cu2+] 0.05911 E=0.34- -log 2 0.01 E=0.34-0.02955xl0g(100) log(100)=2 →E=0.34-0.02955×2=0.34-0.0591 E =0.2809V 20651m Answer: 0.2809 V (e) What is the electrode potential of Cl/Cl half cell at 25°C, if concentration of Cl is 0.05 mol dm3 Cl, + 2é ==2CE E°1.36V Ans. Use Nernst equation Given: Half-cell Cl, + 2é ==2Cl Standard electrode potential (E°) = 1.36V [Ce] = 0.05 mol dm3, T = 25°C = 298K n = 2 = No. of electron transferred 0.0591 E=E° At 25°C (Nernst equation) 1 0.0591 E = 1.36 - -log 2 (0.05)'. 1 E = 1.36 - 0.02955 log 0.0025 E = 1.36 - 0.2955 × 2.60 E = 1.36 - 0.0768 E = 1.28V Answer The electrode potential of Clz/Cl half cell decreases from standard 1.36 to 1.28V, as the [Ce] is decreased to 005M PHOTOVOLTAIC CELLS

The word 'photovoltaic' is composed of photo (light) and volt (electrical potential). The cell which converts light energy into electrical energy is called photovoltaic cell. Principle of photovoltaic cells Definition: A photovoltaic (PV) cell is a device that converts light energy directly into electrical energy using the photovoltaic effeçt. Photovoltaic Effect • The photovoltaic effect is the basic principle behind the working of solar cells. • It is the generation of voltage and current in a material when exposed to light. Working 1. Light Absorption • When sunlight (photons) strikes the surface of a semiconductor, it transfers energy to electrons. 2. Excitation of Electrons • The energy from photons excites electrons, allowing them to jump from the valence band to the conduction band. • This creates free electrons (negative charge) and holes (positive charge). 3. PN Junction Role • A PN junction (junction between p-type and n-type semiconductors) is built inside the PV cell. • The junction helps in separating electrons and holes, preventing recombination. 4. Electric Field Generation • The electric field at the PN junction drives electrons toward the n-side and holes toward the p-side. This movement creates a potential difference (voltage). 5. Current Flow • When the external circuit is connected, electrons flow through the circuit, producing electric current. Merits of photovoltaic cell as sustainable source of energy • The photovoltaic effect continues as long as light is present, making it a sustainable method for generating electricity • A photovoltaic system is a renewable energy source that converts sunlight into electrical energy • The photovoltaic systems use the sun's energy, making them a sustainable energy source that is independent of fossil fuels. WINKLER METHOD, BOD AND DO

Biochemical Oxygen Demand (BOD) • Definition: BOD is the amount of oxygen consumed by microorganisms to decompose organic matter in water over a period of 5 days at 25°C • Significance: It reflects the level of organic pollution in water-higher BOD means more pollution. • Measurement: Calculated by measuring the difference in dissolved oxygen (DO) in a water sample at the start and after 5 days. DO Concentraction • BOD Direction of Flow of River - Fig: Inverse relationship between BOD and DO Winkler Method for DO Measurement Principle: DO in the water oxidizes potassium iodide (KI) to release iodine. Process (i) Add MnSO4 and alkaline KI to the sample. (II) Oxygen reacts, forming a brown precipitate of MnO(OH)2. (111) Acidity the sample → iodine is released (iv) Titrate iodine with sodium thiosulfate using starch as an indicator (blue color indicates presence of iodine). The blue color disappears-this marks the amount of DO. More the amount of iodine formed, more the value of DO. SAMPLE PROBLEMS Sample problem 10.1 See on page number 332 Sample problem 10.2 See on page number 332 Sample problem 10.3 Caculate the oxidation number (Ox. NO.) of manganese in KMnO4. Solution (Ox. NO. of K) + (Ox. No. of Mn) + 4 (Ox. NO. of 0) = 0 + 1 + Mn - 8 = 0 Or Mn = + 7 Thus, the oxidation number (Ox. NO.) of Mn in KMnO4 is + 7. Sample problem 10.4 See on page number 336 Sample problems 10.5 Calculate the mass of lead (Pb) deposited at the cathode during electrolysis when a current of 1:50 A lows through molten lead (II) bromide (PbBr2) for 20.0 min. (relative atomic mass, Ar value: [Pb] = 207; F = 96500 C mol-1) Solution: Step 1: Write the half-equation for the reaction. Step 2: Find the number of coulombs required to deposit 1 mole of product at the electrode. 2 moles of electrons are required per mole of Pb formed Step 3: Calculate the mass by simple proportion using the relative atomic mass. Step 4: Calculate the mass b simple proportion using the relative atomic mass. 193000 C deposits 1 mole Pb, which is 207g Pb So, 1800 C deposits = 1.93 g Pb Sample problem 10.6 What is the electrode potential of a Zn electrode department in a solution containing Zn* ions (0.1M), the e value of Zn*t | ZnO is -0.76V Using Half equation and E°, Zn(ag) + 2e → Z(s) E = (-0.76)+. E = -0.76 + (-0.029) = -0.78V Since, concentration of Zn* is decreased to shift equilibrium backward and electrode potential decreases to -0.78V : . = 2 × 96500 = 193000 C mol = I x t = 1.50 × 20 × 60 = 1800 C E° = - 0.76V 8.31×298 - (n(0.1) 2x 96500