Calculate the emf of the following cell at 298 K if $E_{cell}^o= 1.10 V$:

$Ni(s) + 2Ag^+(0.2M) → Ni^{2+} (0.04M) + 2Ag(s)$

1.10 V

The correct answer is Option (1) → 1.10 V

Use the Nernst equation at 298 K:

$E = E^\circ - \frac{0.0591}{n}\log Q$

Given:

Reaction:

$\text{Ni(s)} + 2\text{Ag}^+ \rightarrow \text{Ni}^{2+} + 2\text{Ag(s)}$

• $E^\circ_{\text{cell}} = 1.10\ \text{V}$
• $n = 2$
• $[\text{Ag}^+] = 0.2\ \text{M}$
• $[\text{Ni}^{2+}] = 0.04\ \text{M}$

Step 1: Reaction quotient Q

$Q = \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} = \frac{0.04}{(0.2)^2} = \frac{0.04}{0.04} = 1$

Step 2: Calculate emf

$E = 1.10 - \frac{0.0591}{2}\log(1)$

Since $\log 1 = 0$,

$E = 1.10\ \text{V}$