Question

What is the value of $E_{\text{cell }}$ ? $Cr\left|C{r}^{3+}(0.1M)\right|\left|F{e}^{2+}(0.01M)\right|Fe$ Given, $E_{C{r}^{3+}/Cr}^{\circ }=-0.74V$ and $E_{F{e}^{2+}/Fe}=-0.44V$

• A. + 0.2941 V
• B. +0.5212 V
• C. +0.1308 V
• D. -0.2606 V

Answer 1

Answer: (A)

$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.059}{n}\log \left[\frac{\text{ product }}{\text{ reactant }}\right]$
Given, $E_{C{r}^{3+}/Cr}^{\circ }=-0.74V$
$E_{F{e}^{2+}/Fe}=-0.44V$
$Cr\left|C{r}^{3+}(0.1M)\right|\left|F{e}^{2+}(0.1M)\right|Fe$
$\therefore C{r}^{3+}/Cr$ is anode and $F{e}^{2+}/Fe$ is cathode.
$E_{\text{cell }}^{\circ }=E_C^{\circ }-E_A^{\circ }$
$=(-0.44)-(0.74)$
$=-0.44+0.74$
$=+0.30V$
Cell reaction is
$2Cr+3F{e}^{2+}⟶2C{r}^{3+}+3Fe$
number of electrons in cell reaction $=6$
$E_{\text{cell }}^{\circ }=E_{\text{cell }}^{\circ }-\frac{0.059}{n}\log \left[\frac{\text{ product }}{\text{ reactant }}\right]$
$=+0.30V-\frac{0.059}{6}\log \left[\frac{{\left(C{r}^{3+}\right)}^2}{{\left(F{e}^{2+}\right)}^3}\right]$
$=0.30-\frac{0.059}{6}\log \left[\frac{(0.1{)}^2}{(0.01{)}^3}\right]$
$=0.30-\frac{0.059}{6}\log 10^4$
$=0.30-\frac{0.059}{6}\times 0.60$
$=0.30-5.9\times 10^{-3}$
$=0.2941V$