Question

At $298$K, the equilibrium constant for the reaction
$Z{n}^{2+}+4N{H}_3\rightleftharpoons \left[Zn\right(N{H}_3{)}_4{]}^{2+}$ is ${10}^9$.
If $E_{\left[Zn\right(N{H}_3{)}_4{]}^{2+}/[Z{n}^{2+}+4N{H}_3]}^o=-1.03$V then the value of $E_{Zn/Z{n}^{2+}}^o$ will be:

• A. $-0.7645$V
• B. $-1.1$V
• C. $+1.1$V
• D. none of these

Answer 1

The correct option is A $-0.7645$V

$\begin{array}{c}\text{Given,}\\ 2n^{+2}+4{NH}_3\rightleftharpoons {\left[Zn{\left({NH}_3\right)}_4\right]}^{+2}({ke}_2={10}^9)\\ \text{We know,}\log k_{eq}=\frac{n{E}^{\circ }}{0.059}(\because T=298k)\\ \Rightarrow \log \left({10}^9\right)=\frac{2\times E^{\circ }}{0.059}\Rightarrow E^{\circ }=0.2655V\end{array}$

$\text{Now,}{E_{\text{cell}}^0=E_{{Zn}^{+2}/Zn}^{\circ }-E_{[Zn}^{\circ }{\left({NH}_3\right)}_4]}^{+2}/Zn$

$Rightarrow0.2655=E_{2n+2/2n}^0-(-1.03)$

$\therefore E_{2n+2/zn}^c=-0.7645v$

so, option(A) is correct.