Question

A cell is given as $M(s) | M^{n+}(aq) || M^{z+}(aq) | M(s)$. For which of the following condition, $E_{\text{cell}}$ is positive:

• A. $C_1<C_2$ (If $C_1$ is concentration at cathode)
• B. $C_2<C_1$ (If $C_1$ is concentration at anode)
• C. $C_1<C_2$ (If $C_2$ is concentration at anode)
• D. $C_1>C_2$ (If $C_1$ is concentration at cathode)

Hint:

In a concentration cell, the spontaneous direction of reaction moves ions from the higher concentration compartment (cathode) to the lower concentration compartment (anode), thereby increasing entropy.

Answer 1

The Correct Option is D

The cell is a concentration cell ($E_{\text{cell}}^o = 0$). We assume $n=z$.
The Nernst equation is: $E_{\text{cell}} = - \frac{0.059}{n} \log \frac{[M^{n+}]_{\text{anode}}}{[M^{n+}]_{\text{cathode}}}$.
For $E_{\text{cell}}$ to be positive, $\log \frac{[M^{n+}]_{\text{anode}}}{[M^{n+}]_{\text{cathode}}}$ must be negative.
This requires the ratio $\frac{[M^{n+}]_{\text{anode}}}{[M^{n+}]_{\text{cathode}}}<1$.
Thus, the condition for spontaneity is $C_{\text{anode}}<C_{\text{cathode}}$.
We analyze Option (D): $C_1$ is the cathode concentration ($C_C$), and $C_2$ is the anode concentration ($C_A$).
$C_1>C_2 \implies C_{\text{cathode}}>C_{\text{anode}}$.
This condition satisfies $C_{\text{anode}}<C_{\text{cathode}}$, making $E_{\text{cell}}$ positive.