Question

E° of the given cell is 1.1 V
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
ΔG° of the cell reaction is
–1.1 F
–2.2 F
–4.4 F
–3.3 F

Answer 1

Answer:

2

Explanation:

G=nFE=2*1.1 F= 2.2 F

n: no of electrons taking part in reaction

F:1 faraday

E: Electrode potential of cell

Answer 2

Given info : E° of the given cell is 1.1 V

                                  Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

To find : ΔG° of the cell reaction is

1. –1.1 F
2. –2.2 F
3. –4.4 F
4. –3.3 F

solution : given, Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

means, Zn + Cu²⁺ ⇄  Zn²⁺ + Cu

no of electrons exchanged from Zn to Cu is 2

n = 2 ...(1)

given standard potential of cell , E° = 1.1 v

using the formula, ΔG° = -nFE°

                                     = -2 × F × 1.1 v = -2.2F

therefore the Gibbs free energy of the reaction is -2.2F