Question

From the following data at $25^{\circ }C$ ,
$C{r}_{(aq)}^{3+}+e^{-}\to C{r}_{(aq)}^{2+},E^{\circ }=-0.424V$
$C{r}_{(aq)}^{2+}+2e^{-}\to C{r}_{(s)},E^{\circ }=-0.900V$
Find $E^{\circ }$ at $25^{\circ }C$ for the reaction,
$C{r}^{3+}+3e^{-}\to C{r}_{(s)}$

• A. - 0.741 V
• B. - 1.324 V
• C. - 0.476 V
• D. + 0.741 V

Answer 1

Answer: (B)

$(i)C{r}^{3+}+e^{-}\to C{r}^{2+},E_1^{\circ }=-0.424V$
$(ii)C{r}^{2+}+2e^{-}\to C{r}^{-},E_2^{\circ }=-9.00V$
$(iii)C{r}^{3+}+3e^{-}\to Cr,E^{\circ }=?$
Since, reaction $(i)$ and $(ii)$ are performed at cathode. Therefore, for reaction $(iii)$
$E^{\circ }=E_1^{\circ }+E_2^{\circ }$
$=(-0.424)+(-0.900)$
$=-1.324V$