Calculate the emf of the following cell : Ni(s)+2Ag^+(0.01 M) → Ni^2+(0.1M)+ 2Ag(s)

Question

Calculate the emf of the following cell :

\(\text{Ni(s)} + 2\text{Ag}^+(0.01\text{M}) \rightarrow \text{Ni}^{2+}(0.1 \text{M}) +2\text{Ag(s)}\)

\(\text{Given that} \ \text{ E}_{cell}^{\circ} = 1.05\text{V}, \ \text{log 10} = 1\)

Answer 1

For the given cell

\(\text{Ni(s)} + 2\text{Ag}^+(0.01\text{M}) \rightarrow \text{Ni}^{2+}(0.1 \text{M}) +2\text{Ag(s)}\)

\(\text{E}_\text{cell} = \text{E} _{\text{cell}}^{\circ} - \frac{0.059}{2}\text{log} \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}\)

\(= 1.05 - \frac{0.059}{2} \text{log} \frac{10^{-1}}{(10^{-2})^2}\)

\(= 1.05 - \frac{0.059}{2}\text{log 10}^3\)

\(= 1.05 -0.09\)

\(= 0.96 \text{ volt }\)