Chemistry - Electrochemistry Question with Solution | TestHub

The overall cell reaction is:

$Zn+2H^{+}(g)⟶Z{n}^{2+}+H_2$

$E_{cell}=E_{cell}^{\circ }-\frac{0.059}{2}\log \frac{[Z{n}^{2+}]P_{H_2}}{[H^{+}{]}^2}$

If $E_{cell}=E_{cell}^{\circ }$, then:

$$\begin{gathered} \text{displaylines}\log \frac{[Z{n}^{2+}][H^{+}{]}^2}{[H_{{}_2}]}=0 \\ \frac{[Z{n}^{2+}][H^{+}{]}^{}}{[H_{{}_2}]}=1 \end{gathered}$$