First slide

Nernst Equation

Question

Given $E_{Cr + 3 / Cr}^{0} = - 0.72 V; E_{{Fe}^{2 + / F_{e}}}^{0} = - 0.42 V$
The potential for the cell,
Cr|Cr³⁺(0.1M||Fe²⁺(0.01M| Fe is

Moderate

A

– 0.339 V
B

– 0.26 V
C

0.26 V
D

0.339 V

Solution

From the given SRP values Cr electrode acts as anode Fe electrode acts as cathode.
Cell reaction is $2 {Cr}_{(s)} + 3 {Fe}_{(aq)}^{+ 2} \to 2 {Cr}_{(aq)}^{+ 3} + 3 {Fe}_{(s)}$
Applying Nernst equatioin $E_{cell} = [E_{c}^{o} - E_{a}^{o}] - \frac{0.0591}{6} \log \frac{{[{Cr}^{+ 3}]}^{2}}{{[{Fe}^{+ 2}]}^{3}}$
$E_{cell} = (- 0.42 - (- 0.72)) - \frac{0.0591}{6} \log \frac{[0.1]^{2}}{[0.01]^{3}}$
$E_{cell} = 0.3 - \frac{0.059}{6} \log 10^{4} = 0.26 V$

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