Question

Construct the cell corresponding to the reaction:
$3C{r}^{2+}(aq,1M)\to 2C{r}^{3+}(aq,1M)+Cr\left(s\right)$ and predict if the reaction is spontaneous or not. Also calculate $\Delta$H and $\Delta$S of the reaction at ${25}^{\circ }$C.
Given: $E_{C{r}^{3+}\left(aq\right)/Cr\left(aq\right)}=0.5V{E}_{C{r}^{3+}\left(aq\right)/C{r}^{2+}\left(aq\right)}=-0.41V$
$\Delta$G of the reaction at ${35}^{\circ }C=-270.50kJ$

• A. Non-spontaneous
  $\Delta H=-51.05kJ,\Delta S=700kJ{K}^{-1}$
• B. Non-spontaneous
  $\Delta H=-51.05kJ,\Delta S=700kJ{K}^{-1}$
• C. Spontaneous
  $\Delta H=-51.05kJ,\Delta S=706kJ{K}^{-1}$
• D. Spontaneous
  $\Delta H=-51.05kJ,\Delta S=706kJ{K}^{-1}$

Answer 1

The correct option is D Spontaneous

$\Delta H=-51.05kJ,\Delta S=706kJ{K}^{-1}$
The cell corresponding to the given reaction is as follows:

$C{r}^{2+}(aq,1M),C{r}^{3+}\left(aq1M\right)||C{r}^{2+}\left(aq1M\right)|Cr\left(s\right)$

$I.C{r}^{3+}\left(aq\right)+3e^{-}\to Cr\left(s\right);\Delta G_I^0=-3F{E}_{C{r}^{3+}\left(aq\right)/Cr\left(s\right)}^0$

$II.C{r}^{3+}\left(aq\right)+e^{-}\to C{r}^{2+}\left(aq\right);\Delta G_{II}^0=-F{E}_{C{r}^{3+}\left(aq\right),C{r}^{2+}\left(aq\right)}^0$

$Eq.I-Eq.II,C{r}^{2+}\left(aq\right)+2e^{-}\to Cr\left(s\right);\Delta G^0=(-3\times 0.5F)-0.41F=-1.91F$

$\Delta G^0=-2F{E}_{C{r}^{2+}\left(aq\right)/Cr\left(s\right)}^0=-1.91F$
$\therefore E_{C{r}^{2+}\left(aq\right)/Cr\left(s\right)}^0=\frac{1.91}{2}=0.955V$

$E_{Cell}^0=E_{C{r}^{2+}\left(aq\right)/Cr\left(s\right)}^0-E_{C{r}^{3+}\left(aq\right),C{r}^{2+}\left(aq\right)}^0=0.955+0.41=1.365V$

$\Delta G^0=-nF{E}_{Cell}^0=-2\times 96500\times 1.365J=-263.44kJ$
Since, $\Delta G^0$ is $-ve$ hence the given reaction is spontaneous.

From Gibbs-Helmholtz equation,

$\Delta G=\Delta H+T{\left[\frac{\partial \left(\Delta G\right)}{\partial T}\right]}_p$

$-263.44=\Delta H+298\frac{(-270.50+263.44)}{10}=\Delta H-(298\times 0.706)$

$\therefore \Delta H=-53.05kJ$

$\Delta G=\Delta H-T\Delta S$
$\therefore \Delta S=\frac{\Delta H-\Delta G}{T}=\frac{-53.05+263.44}{298}=0.706kJ{K}^{-1}$