Calculate the EMF of the cell $\text{Zn}|{\text{Zn}}^{2+}(0.1\text{ M})||{\text{Ag}}^{+}(0.01\text{ M})|\text{Ag}$ at $298\text{ K}$. Given $E_{{\text{Zn}}^{2+}/\text{Zn}}^{\circ }=-0.76\text{ V}$ and $E_{{\text{Ag}}^{+}/\text{Ag}}^{\circ }=0.80\text{ V}$.

Solution

1. The cell reaction is $\text{Zn}+2{\text{Ag}}^{+}\to {\text{Zn}}^{2+}+2\text{Ag}$ with $n=2$.
2. $$\begin{array}{rl}{E}_{\text{cell}}^{\circ }& =E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }\\ & =0.80-(-0.76)\\ & =1.56\text{ V}\end{array}$$.
3. Using Nernst equation: $E=E^{\circ }-(0.0591/2)\log ([{\text{Zn}}^{2+}]/[{\text{Ag}}^{+}{]}^2)$.
4. $$\begin{array}{rl}E& =1.56-0.02955\log (0.1/(0.01{)}^2)\\ & =1.56-0.02955\log (1000)\end{array}$$.
5. $$\begin{array}{rl}E& =1.56-0.02955\times 3\\ & =1.56-0.08865\\ & =1.47135\text{ V}\end{array}$$.
6. Hence the answer is (A).