Question

If $E_{\text{cell }}^{\circ }$ is $1.05V$, the emf of the cell for the following cell reaction
$Ni(s)+2A{g}^{+}(0.004M)⟶N{i}^{2+}(0.16M)+2Ag(s)$ at $298K$ in $V$ is

• A. 0.932
• B. 1.227
• C. 0.732
• D. 1.397

Answer 1

Answer: (A)

Cell reaction

$\underset{(s)}{\mathrm{Ni}}+\underset{(0.004M)}{2A{g}^{+}}⟶\underset{(0.16M)}{N{i}^{2+}}+\underset{(s)}{2Ag}$

Given,

$\left[A{g}^{+}\right]=0.004M$

$\left[N{i}^{2+}\right]=0.16M$

$n=2$

$E_{cell}^{\circ }=1.05V$

Apply in the Nernst equation,

$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{n}\log \frac{\left[N{i}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}$

Put the values in equation

$E_{\text{cell }}=1.05-\frac{0.0591}{2}\log \frac{[0.16]}{[0.004{]}^2}$

$E_{\text{cell }}=1.05-\frac{0.0591}{2}\log \frac{16}{16}\times 10^4$

$E_{\text{cell }}=1.05-\frac{0.059}{2}\log 10^4$

$E_{\text{cell }}=1.05-\frac{0.059}{2}4\log 10[\because \log 10=1]$

$E_{\text{cell }}=1.05-\frac{0.059}{2}\times 4$

$E_{\text{cell }}=1.05-0.118\Rightarrow E_{\text{cell }}=0.932V$