Question

Given, $E^{\circ} _{Cr^{3+}/Cr} = -0.72 V, E^{\circ}_{Fe^{2+}/Fe} = -0.42 V$ .The potential for the cell
$r \left|Cr^{3+}\left(0.1 \,M\right)\right|\left|Fe^{2+} \left(0.01 \,M\right)\right|Fe $ is

• A. $0.26\, V$
• B. $0.399\, V$
• C. $ - 0.399\, V$
• D. $ - 0.26\, V$

Answer 1

Answer: (A)

$Cr \left| Cr ^{3+}(0.1 M ) \| Fe ^{2+}(0.01 M )\right| Fe$

Oxidation half-cell

$\left. Cr \longrightarrow Cr ^{3+}+3 e^{-}\right] \times 2$

Reduction half-cell

$\left. Fe ^{2+}+2 e^{-} \longrightarrow Fe \right] \times 3$

Net cell reaction

$2 Cr +3 Fe ^{2+} \longrightarrow 2 Cr ^{3+}+3 Fe , n=6$

$ E_{\text {cell }}^{\circ} =E_{\text {oxi }}^{\circ}+E_{\text {red }}^{\circ} $

$=0.72-0.42=0.30 \,V $

$ E_{\text {cell }} =E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{\left[ Cr ^{3+}\right]^{2}}{\left[ Fe ^{2+}\right]^{3}} $

$=0.30-\frac{0.0591}{6} \log \frac{(0.1)^{2}}{(0.01)^{3}} $

$=0.30-\frac{0.0591}{6} \log \frac{10^{-2}}{10^{-6}} $

$=0.30-\frac{0.0591}{6} \log 10^{4}$

$E_{\text {cell }} =0.2606\, V $