Consider the above electrochemical cell where a metal electrode ( $M$ ) is undergoing redox reaction by forming $\mathrm{M}^{+}\left(\mathrm{M} \rightarrow \mathrm{M}^{+}+\mathrm{e}^{-}\right)$. The cation $\mathrm{M}^{+}$is present in two different concentrations $c_1$ and $c_2$ as shown above. Which of the following statement is correct for generating a positive cell potential?
Select the correct option:
A
If $c_1$ is present at anode, then $c_1=c_2$
B
If $\mathrm{c}_1$ is present at cathode, then $\mathrm{c}_1<\mathrm{c}_2$
C
If $c_1$ is present at cathode, then $c_1>c_2$
D
If $c_1$ is present at anode, then $c_1>c_2$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
(1) If $\mathrm{C}_1$ is at anode ⇒ cell reaction
$$
\begin{aligned}
& \mathrm{M}^{+}\left(\mathrm{C}_2\right) \rightarrow \mathrm{M}^{+}\left(\mathrm{C}_1\right) \\
& \mathrm{E}_{\text {cell }}=-0.059 \log \frac{\mathrm{C}_1}{\mathrm{C}_2} \\
& \therefore \mathrm{E}_{\text {cell }}>0 \Rightarrow \mathrm{C}_1<\mathrm{C}_2 \end{aligned} $$ (2) If $\mathrm{C}_1$ is at cathode $$ \begin{aligned} & \mathrm{M}^{+}\left(\mathrm{C}_1\right) \rightarrow \mathrm{M}^{+}\left(\mathrm{C}_2\right) \\ & \mathrm{E}_{\text {cell }}=-0.059 \log \frac{\mathrm{C}_2}{\mathrm{C}_1}>0 \\
& \mathrm{C}_2<\mathrm{C}_1
\end{aligned}
$$
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