Question

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

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

Answer 1

Answer: (A)

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