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For the following cell,

$Zn(s)\mid ZnSo_{4}(aq)\parallel CuSo_{4}(aq)\mid Cu(s)$

when the concentration of Zn²⁺ is 10 times the concentration of Cu²⁺, the expression of ΔG (in J mol^-1) is

[ F is faraday constant : R is gas constant : T is temperature :E⁰ (cell) =1.1 V]

A) 2.303 RT + 1.1 F

B) 1.1 F

C) 2.303 RT-2.2 F

D) -2.2 F

Option C

The redox reaction is

$Zn(s)+Cu^{2+}\rightarrow Zn^{2+}+Cu$

The nernst equation is

$E= E^{0}-\frac{2.303RT}{2F}log10$

$=1.1-\frac{2.303 RT}{2F}$

Also, $\triangle G=-nEF =-2F(1.1-\frac{2.303RT}{2F})$

=-2.2F +2.303RT

=2.303RT-2.2F

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