For the following cell, $$\mathrm{Zn}(s)\left|\mathrm{ZnSO}_{4}(a q) \| \mathrm{CuSO}_{4}(a q)\right| \mathrm{Cu}(s)$$ when the concentration of $\mathrm{Zn}^{2+}$ is 10 times the concentration of $\mathrm{Cu}^{2+}$, the expression for $\Delta G$ (in $\mathrm{J} \mathrm{mol}^{-1}$) is [F is Faraday constant; R is gas constant; T is temperature; $E^{o}($ cell $)=1.1 \mathrm{~V}$]

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Redox Reactions and Electrochemistry

For the following cell,

Zn (s) ∣ ZnSO_{4} (a q) ∥ CuSO_{4} (a q) ∣ Cu (s)

when the concentration of

Zn^{2 +}

is 10 times the concentration of

Cu^{2 +}

, the expression for

Δ G

(in

J mol^{- 1}

) is
[F is Faraday constant; R is gas constant; T is temperature;

E^{o} (

cell

) = 1.1 V

]

Answer: (B)

For the cell reaction:

Zn + Cu^{2 +} \to Zn^{2 +} + Cu

n = 2

electrons.

Using the Nernst equation:

E_{cell} = E_{cell}^{o} - \frac{2.303 R T}{n F} lo g \frac{[ Zn ^{2 +} ]}{[ Cu ^{2 +} ]}

E_{cell} = 1.1 - \frac{2.303 R T}{2 F} lo g 10 = 1.1 - \frac{2.303 R T}{2 F}

Since

Δ G = - n F E_{cell}

Δ G = - 2 F (1.1 - \frac{2.303 R T}{2 F}) = 2.303 R T - 2.2 F

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option