Question

Scores A Review | Constants | Periodic Table e Text Standard Reduction Half-Cell Potentials at 25°C Assignments Half-Reaction Course Tools + Pb2+ (aq) + 2 H2O(l)| PbO2(s) + 4 H+ (aq) + 2 e Cl2 (9) +2 e +2 CT (aq) Pb2+ (aq) +2 e e → Pb(s) + Pb(s) E° (V) 1.46 1.36 -0.13

Answer 1

According to Nernst equation

AG= -nFE

where n = number of electrons, F is Faraday constant and E is the voltage

The unit of Faraday constant is C/mol (coulomb/mol) and E is V (volt)

AG = (C/mol) x V = cx molJ/mol

PbO₂ + 4 H⁺ (aq) + 2e^-$\rightarrow$ Pb²⁺(aq) + 2 H₂O eq. 1.1   

AG = -2 x (+1.46) F

Pb²⁺ (aq) + 2 e^-$\rightarrow$ Pb(S) eq 2.2   

AG = -2x (-0.13) F

Taking sum of the equations 1.1 and 1.2 we get

PbO₂ + 4 H⁺ (aq) + Pb²⁺ (aq) + 2 e^- + 2e^-$\rightarrow$ Pb²⁺(aq) + Pb(S) + 2 H₂O eq 1.3  

AG = -2 x (+1.46 – 0.13) F

The reduction reaction is equation 1.4 as follows

PbO₂ + 4 H⁺ (aq) + 4e^-$\rightarrow$  Pb(S) + 2 H₂O eq 1.4

AG = -2 x (+1.33) F

The E⁰red for the reduction is = $\frac{\Delta G}{-4F}$ = +0.66 V

Cl₂ (g) + 2e^-$\rightarrow$ 2 Cl^- (aq) eq 1.5 E⁰_red = +1.36 V

The equation for the oxidation is as follows (eq 1.6)

2 Cl^- (aq) $\rightarrow$ Cl₂(g) + 2e^- eq 1.6 E⁰_ox = -1.36 V

AG = -2 x (+1.36) F

The number of electrons for the reduction must be same for the oxidation. Thus,

4 Cl^- (aq) $\rightarrow$ 2 Cl₂(g) + 4e^- eq 1.7 E⁰_ox = -1.36 V

AG = -4 (-1.36) F

Taking sum of the equations 1.4 and 1.7;

PbO₂ + 4 H⁺ (aq) + 4e^-$\rightarrow$  Pb(S) + 2 H₂O eq 1.4 E0_red = +0.66 V   

AG = -4x (+0.66) F

4 Cl^- (aq) $\rightarrow$ 2 Cl₂(g) + 4e^- eq 1.7 E⁰_ox = -1.36 V

AG = -4 (-1.36) F

we get equation 1.8 (answer to part B);

PbO₂ + 4 H⁺ + 4 Cl^- (aq) $\rightarrow$ Pb(S) + 2 H₂O + 2 Cl₂(g)   

AG = -4x (+0.66 – 1.36)F = -4 x 0.70F

Thus E⁰_cell = = +0.70 V (answer to part C)

AG) – 4F