Question

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

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

Answer 1

Answer: (A)

$Cr\left|C{r}^{3+}(0.1M)\Vert F{e}^{2+}(0.01M)\right|Fe$

Oxidation half-cell

$Cr⟶C{r}^{3+}+3e^{-}]\times 2$

Reduction half-cell

$F{e}^{2+}+2e^{-}⟶Fe]\times 3$

Net cell reaction

$2Cr+3F{e}^{2+}⟶2C{r}^{3+}+3Fe,n=6$

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

$=0.72-0.42=0.30V$

$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{n}\log \frac{{\left[C{r}^{3+}\right]}^2}{{\left[F{e}^{2+}\right]}^3}$

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

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

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

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