Question

Given $E^{o}_{Cr^{3+}/Cr} = -0.72 V, E^{o}_{Fe^{2+}/Fe} =-0.42 V.$ The poten tial for the cell
$Cr | Cr^{3+} (0.1 M) || Fe^{2+} (0.01 M)| Fe$ is

• A. $0.26$ V
• B. $0.339$ V
• C. $-0.399$ V
• D. $-0.26$ V

Answer 1

Answer: (A)

$As E^{o}_{Cr^{3+}Cr}=-0.7 V$ and $E^{o}_{Fe^{2+}/Fe}=-0.42 V $
$2Cr +3Fe^{2+} \to3Fe+2Cr^{3+} $
$E_{cell}=E^{o}_{cell}-\frac{0.0591}{6}log \frac{\left[Cr^{3+}\right]^{2}}{\left[Fe^{2+}\right]^{3}} $
$=\left(-0.42 +0.72\right)-\frac{0.0591}{6}log \frac{\left[0.1\right]^{2}}{\left[0.01\right]^{3}}$
$=0.30-\frac{0.0591}{6} log \frac{\left[0.1\right]^{2}}{\left[0.01\right]^{3}}$
$=0.30-\frac{0.0591}{6}log10^{4} $
$E_{cell}=0.2606 \,V$