Chapter 10: Electrochemistry

Electrical double layer is formed at the interface between a metal electrode and an electrolyte solution due to the separation of charges. When a metal is dipped into its salt solution, metal atoms lose electrons and become positive ions, which move into the solution, while the electrons remain on the electrode. This creates a layer of positive ions in the solution and negative charge on the metal surface, forming two layers of hence called the opposite charges electrical double layer Reason: It formed to maintain electrochemical equilibrium between the electrode and the electrolyte.

The electrode potential of copper (Cu) is called reduction potential because it is measured for the reaction in which Cu2+ ions gain electrons to form Cu metal: (u(a) + 2é → CU(s) This process involves the gain of electrons, which is defined as reduction, so the potential is called reduction potential. What are the advantages of salt bridge in a galvanic cell? The advantages of a salt bridge in a galvanic cell are: electrical neutrality by (i) Maintains allowing the flow of ions between the two half-cells. (ii) Completes the electrical circuit to allow continuous flow of electrons (iii) Prevents mixing of different solutions, which could cause direct chemical reaction and stop the cell from working.

See Q15. from theory.

The feasibility of a chemical reaction can be predicted using the cell voltage (E If the E° is cell is positive, the reaction spontaneous and feasible If the E° 'cell is negative, the reaction is non- spontaneous and not feasible. E° 'anode cell = E° cathode - E°

See Q8. from theory NUMERICAL PROBLEMS (EXERCISE)

During electrolysis of aqueous NaCl, Nat ions are not liberated at the cathode because water is reduced more easily than sodium ions. Water has a higher reduction potential than Nat, so H2 gas is produced instead of sodium metal. 2H2O+2é → H 2(g) + 20H

See Q12. from theory is meant by Standard

(i) CrCl Let the oxidation number of Cr = x Chlorine (CI) has an oxidation number of -1 x+3(-1)=0=x-3=0=x=+3 Oxidation number of Cr = +3 (ii) Cr2(SO4)3 Let oxidation number of Cr = x Sulfate ion (SO4 ) has a charge of -2 Total charge from 3 sulfate ions = 3 x (-2) = -6 So for the compound Cr2(SO4)3: 2x+(-6)=0→2x=+6→x=+3 Oxidation number of Cr = +3 (iii) Cr,O: (Dichromate ion) Let oxidation number of Cr = x Oxygen (O) has an oxidation number of -2 Total charge from 7 oxygen atoms = 7 x (-2) =-14 Total charge on. the ion = -2 2x+(-14)=-2 = 2x=+12→x= + 6 Oxidation number of Cr = +6

See Q10. from theory. Step 3: Plug values into the Nerst equation - In 100 E = -0.76- (8.31)(298) (2)(96500) Calculate the constant term 2477.66 (8.31)(298) = = 0.012837V 193000 (2)(96500) Calculate In 100 In 100 = 1n(10) =2 1n 10=2×2.303 = 4.606 Step 4: Calculate the potential E = - 0.76 - (0.012837 × 4.606) Answer: E= -0.82V

(i) Copper is added to a solution of magnesium sulfate (MgSO4): Reaction prediction: Since copper (Cu) is less reactive than magnesium (Mg), no reaction will occur No reaction, because copper cannot displace magnesium from its salt solution. (ii) Iron is added to a dilute solution of hydrochloric acid (Hcl): Reaction prediction: Iron (Fe) is more reactive than hydrogen, so it will react with HCl to produce hydrogen gas. Balanced equation: Fes) + 2HCC,

Metals higher in the activity series are more reactive and can easily lose electrons (oxidize). When these metals react with acids, they displace hydrogen ions (H) from the acid because they have a stronger tendency to oxidize compared to hydrogen. On the other hand, metals lower in the activity series have a weaker tendency to lose electrons and cannot displace hydrogen ions from the acid. These metals are less reactive, and this hydrogen is not displaced in reactions with acids. In simple terms • Higher activity series metals (e.g., Mg, < Zn): More reactive → can displace H 10ns. • Lower activity series metals (e.g., Cu, Pt): Less reactive → cannot displace H+ ions

wer: 1.18g of copper is deposited at the cathode

To calculate the of Faradays required to deposit the given , we use masses of Ag+, Cu2+, and Al3+ Faraday's Laws of Electrolysis. The formula to calculate the number of Faradays is: Mass of substance Faradays = Molar mass of substance x n Where • n is the number of electrons involved in the reduction process for each ion. • Molar mass is the molar mass of the substance. (i) For Ag* (Silver ion) • Molar mass of Ag = 108 g/mol • Ag+ gains 1 electron to form Ag (s), so n = 1 108 g Faradays for Ag =- - =1 Faradays 1 mol-1 (ii) For Cu? (Copper ion) Molar mass of Cu = 63.5 g/mol • Cu gains 2 electrons to form Cu (s), so n = 2 Faradays for 63.5 63.5 g Cu = 63.5 g / mol× 2 127 = 0.5 Faraday (ill) For Al* (Aluminum ion) • Molar mass of Al = 27 g/mol • Al' gains 3 electrons to form Al (s), so n = 3 27 27g Faradays for Al = = 0.333 Faradays 27g/ mol× 381 The total number of Faradays required to , and 27 deposit 108 g of Ag, 63.5 g of Cu+ g of Al'* is 1.833 Faradays QI0. In an electrolysis experiment, a current of 0.500 A was passed through a solution of AgNO3 for 30.0 minutes. The mass of silver deposited on the cathode was found to be 0.503 g. Given that the molar mass of silver is 107.87 g mol and the charge on a silver ion is +1. Calculate the value of Avogadro's number (NA) from this data. Ans. To calculate Avogadro's number (NA) from electrolysis data, we use Faraday's Laws of Electrolysis and relate the total charge passed to the number of atoms deposited. Given Data: • Current (I) = 0.500 A • Time (t) = 30.0 minutes = 30 × 60 = 1800 seconds • Mass of silver deposited (m) = 0.503 g • Molar mass of Ag (M) = 107.87 g/mol • Charge on Ag ion = +1 = 1 mole of Ag* needs 1 mole of electrons (1 Faraday = 96485 C) Step 1: Calculate total charge (Q) passed Q= Ix t= 0.500 A x 1800 s = 900 C Step 2: Calculate moles of Ag deposited: Mass Moles of Ag= Molar Mass 0.503 Moles of Ag = 107.87 Moles of Ag= 4.664× 10- mol Step 3: Calculate moles of electrons used Since 1 mole of Ag+ requires 1 mole of electrons: Moles of electrons = Moles of Ag = 4.664 × 10-3 mol Step 4: Use charge to find charge per mole (Faraday constant) Charge per moles (F) = Moles of Ag 900 = 4.664x10-3 ee 1. F = 1.93 x C/mol Step 5: Use Faraday's constant to find Avogadro's number Faraday's constant (F) = NAx e e = 1.602 × 10-19 C (charge of one electron) F 1.93x105 NA===. e 1.66.02 × 10-19=6.6.02 × 10^23

wer for (a): Ni (s) + 2H (b) Determine the standard electrode potential (E°) of the N'/Ni half-cell. Given: • Cell potential Ecell = 0.25 V • SHE is always assigned E° = 0.00 V • Nickel is the anode, so: E°cell = E°cathodé - E° Substitute values: 0.25 = 0.00 - E° Solve for 0.25 = 0.00 - E° N?* /N; E° N IN, = - 0.25V Answer for (b): E® = -0.25V N?* /N. (c) Identify which electrode is the anode and which is the cathode. From the problem: • Nickel electrode is negative terminal → anode • SHE is positive terminal →> cathode Answer for (c): Anode : Ni) Cathode: SHE(H, / Ht) (aq) / Nts) halt-cell. (E° = 0.00 V) anode N?+ /N;

wer. (ii) Write the balanced overall cell reaction. the (iii) Calculate standard cell potential (E° 'cell). Ans. Given Standard Electrode Potentials: • Ni(ag) + 2e = = Ni (s) E° = -0.25 V E° = - 0.28 V (i) Identify anode and cathode: • In a galvanic (voltaic) cell, oxidation occurs at the anode and reduction at the cathode. • The electrode with lower (more negative) reduction potential is more likely to oxidize (lose electrons). • Here, Co has a lower E° (-0.28 V) than Ni (-0.25 V), so Co is oxidized and acts as the anode. Answer • Anode: Cobalt (Co) • Cathode: Nickel (Ni) Justification: Co has a lower (more negative) standard SLO BASED SHORT QUESTION ANSWERS

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It includes the study of redox reactions where electrons are transferred. Example: Electrolysis of water produces hydrogen and oxygen gases using electricity.

Photosynthesis is a redox reaction which provides food for the entire planet, and another one is respiration that keeps us alive, both are redox reaction

A device that converts chemical energy into electrical energy through redox reactions. Example: Daniell cell (Zn-Cu cell). 018. What is the function of a salt bridge in a galvanic cell? Ans. A salt bridge maintains electrical neutrality by allowing the movement of ions between the two half-cells. Example: A KNOs salt bridge connects the Zn and Cu compartments in a Daniell cell.

Conductors (usually metals) where oxidation or reduction occurs. • Anode: where oxidation occurs • Cathode: where reduction occur

A cell that generates electrical energy reaction. from a spontaneous redox Example: Daniell cell: ZM (s) / Zn(ag) | Cu(ag) / Cu(s)

The anode is negative because it loses electrons (oxidation)

from Electrons flow anode be used in cathode. From where it can external circuit.

It is a classic example of a galvanic cell that demonstrates conversion of chemical to electrical energy

A device where electrical energy is used to drive a non-spontaneous chemical reaction. Example: Electrolysis of molten NaCl.

Anode is positive because it attracts negative ions (oxidation still occurs here).

The process of depositing a metal layer on an object using an electrolytic cell. Example: Electroplating silver on a spoon

The potential difference between a metal electrode and its ion solution. is standard electrode

The electrode potential measured under standard conditions: 1M, 1 atm, 25°C.

Because it is stable, reversible, and easy to reproduce. Cell Potential and EMF

Anode/Anode sol. II Cathode sol. / Cathode Example: ZM (s) / Zn(a (ag) | Cu(ag) / Cu(s)

A list of elements arranged in order of increasing standard electrode potentials. 036. How does the electrochemical series help in predicting displacement reactions? Ans. A metal higher in the series can displace a lower metal from its salt solution. Example: ZIl) + CuSO, →>ZıSO, + Cu(s potassium a strong

Because it has a very negative E° value (-2.92 V), indicating high tendency to lose electrons. Applications

In batteries such as dry cells, lead-acid batteries, and lithium-ion cells.

A galvanic cell using MnOz as cathode and Zn as anode with an electrolyte paste. corrosion in terms of 040. What is electrochemistry? Ans. A redox process where metals (like iron) get oxidized due to environmental exposure Example: Miscellaneous Concepts

The mass of substance deposited or liberated is directly proportional to the quantity of electricity passed. number of

(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. (b) If an element in a species undergoes both increase in oxidation number or in decrease number oxidation 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 for increase in the equation. First oxidation number while second is written for decrease in oxidation number Cf2 + NaOH → NaC/ + H2O+ NaOC( (c) If an element in a species undergoes increase or decrease in oxidation number 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(NO.): + H2O+ NaOC: HC(+ K,CrO, + HC →C + Kcl+ Cr, + H.O

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, DESCRIPTIVE QUESTIONS (EXERCISE)