Question

Calculate the equilibrium constant for the redox reaction at ${25}^o$C, $S{r}_{\left(s\right)}+M{g}_{\left(aq\right)}^{2+}\to S{r}_{\left(aq\right)}^{2+}+M{g}_{\left(s\right)}$, that occurs in a galvanic cell. $E_{M{g}^{2+}/Mg}^o=-2.37$V and $E_{{Sr}^2+/Sr}^o=-2.89$V.

• A. $2.69\times {10}^{15}$
• B. $2.69\times {10}^{17}$
• C. $3.69\times {10}^{17}$
• D. $3.69\times {10}^{15}$

Answer 1

The correct option is C $3.69\times {10}^{17}$

(i) ${M{g}^{2+}}_{\left(aq\right)}+2e^{-}⟶M{g}_{\left(s\right)};E_{M{g}^{2+}/Mg}^o=-2.37V$

${S{r}^{2+}}_{\left(aq\right)}+2e^{-}⟶S{r}_{\left(s\right)};E_{S{r}^{2+}/Sr}^o=-2.89V$

(ii) $S{r}_{\left(s\right)}⟶S{r^{2+}}_{\left(aq\right)}+2e^{-};E_{Sr/S{r}^{2+}}^o=2.89V$

Adding (i) & (ii), we get,

${Sr}_{\left(s\right)}+M{g}_{\left(aq\right)}^{2+}⟶S{r}_{\left(aq\right)}^{2+}+M{g}_{\left(s\right)};E^o=-2.37+2.89=0.52V$

Now, $\Delta G^o=-RTln{K}_{eq}$

$\therefore -nF{E}^o=-RTln{K}_{eq}$

$\therefore ln{K}_{eq}=\frac{nF{E}^o}{RT}=\frac{2\times 96500\times 0.52}{8.314\times 298}$

$\therefore ln{K}_{eq}=40.507$

$\therefore K_{eq}=3.69\times {10}^{17}$