Question

Calculate the emf of the cell in which of the following reaction takes place:
$\text{Ni}\left( \text{s} \right)+2\text{A}{{\text{g}}^{+}}\left( 0.002\text{M} \right)\xrightarrow{{}}\text{N}{{\text{i}}^{2+}}\left( 0.160\text{M} \right)+2\text{Ag}\left( \text{s} \right)$
Given that ${{\text{E}}^{\circ }}=1.05\ \text{V}$
A. $0.73\ \text{V}$
B. $0.91\ \text{V}$
C. $0.62\ \text{V}$
D. $0.34\ \text{V}$

Answer 1

Hint: The emf of the cell can be calculated using Nernst equation. Determine the number of electrons transferred and the reaction constant is determined as concentrations of products divided by the concentrations of reactants.
Formula Used:
${{\text{E}}_{\text{cell}}}={{\text{E}}^{\circ }}-\dfrac{0.0591}{n}\log {{Q}_{C}}$

Complete answer:
Nernst equation is used to determine the cell potential during the non-standard state. It can be mathematically written as,
${{\text{E}}_{\text{cell}}}={{\text{E}}^{\circ }}-\dfrac{0.0591}{n}\log {{Q}_{C}}$
Here, ${{\text{E}}_{\text{cell}}}$ is the cell potential, ${{\text{E}}^{\circ }}$ is the cell potential at standard state conditions, $n$ is the number of electrons transferred and ${{Q}_{C}}$ is the reaction quotient.
The cell reaction given is,
$\text{Ni}\left( \text{s} \right)+2\text{A}{{\text{g}}^{+}}\left( 0.002\text{M} \right)\xrightarrow{{}}\text{N}{{\text{i}}^{2+}}\left( 0.160\text{M} \right)+2\text{Ag}\left( \text{s} \right)$
From the above cell reaction, we can see that 2 electrons are transferred. Thus, $n=2$.
The reaction quotient ${{Q}_{C}}$ can be calculated by dividing the concentration of the product with the concentration of reactants. This can be seen as,
${{Q}_{C}}=\dfrac{\left[ \text{N}{{\text{i}}^{2+}} \right]}{{{\left[ \text{A}{{\text{g}}^{+}} \right]}^{2}}}$
Thus, the Nernst equation now becomes,
${{\text{E}}_{\text{cell}}}={{\text{E}}^{\circ }}-\dfrac{0.0591}{n}\log \dfrac{\left[ \text{N}{{\text{i}}^{2+}} \right]}{{{\left[ \text{A}{{\text{g}}^{+}} \right]}^{2}}}$
The cell equation mentioned in the question gives $\left[ \text{N}{{\text{i}}^{2+}} \right]=0.160\text{M}$, $\left[ \text{A}{{\text{g}}^{+}} \right]=0.002\text{M}$ and ${{\text{E}}^{\circ }}=1.05\ \text{V}$ . Now, substitute all the obtained values in the Nernst equation and calculate the cell potential. This gives,
$\begin{align}
  & {{\text{E}}_{\text{cell}}}={{\text{E}}^{\circ }}-\dfrac{0.0591}{n}\log \dfrac{\left[ \text{N}{{\text{i}}^{2+}} \right]}{{{\left[ \text{A}{{\text{g}}^{+}} \right]}^{2}}} \\
 & =1.05-\dfrac{0.0591}{2}\log \dfrac{0.160}{{{\left( 0.002 \right)}^{2}}} \\
 & =1.05-0.02955\log \left( 4\times {{10}^{4}} \right) \\
 & =0.91\ \text{V}
\end{align}$

Thus, the correct option is (B).

Note:
Correctly determine the number of electrons transferred. Correctly determine the value of logarithmic of the reaction quotient as using natural log can lead to different results. Determine the reaction coefficient as it is obtained as the division of concentration of product with the concentration of reactants.