Question

Consider the following electrolytic cells
(i) M(s) | M\textsuperscript{2+(aq), 0.1M || X\textsuperscript{2+}(aq), 0.01M | X(s)
(ii) M(s) | M\textsuperscript{2+}(aq), 0.1M || X\textsuperscript{2+}(aq), 0.1M | X(s) and
(iii) M(s) | M\textsuperscript{2+}(aq), 0.01M || X\textsuperscript{2+}(aq), 0.1M | X(s)
The cell EMF of the above cells are E\textsubscript{1}, E\textsubscript{2} and E\textsubscript{3} respectively. Which one of the following is true?}

• A. E\textsubscript{1} $>$ E\textsubscript{2} $>$ E\textsubscript{3}
• B. E\textsubscript{2} $>$ E\textsubscript{3} $>$ E\textsubscript{1}
• C. E\textsubscript{3} $>$ E\textsubscript{1} $>$ E\textsubscript{2}
• D. E\textsubscript{1} $>$ E\textsubscript{3} $>$ E\textsubscript{2}
• E. E\textsubscript{3} $>$ E\textsubscript{2} $>$ E\textsubscript{1}

Hint:

Think of chemical potential like water pressure: higher concentration at the "source" (Cathode) and lower concentration at the "sink" (Anode) creates the highest flow (EMF)!

Answer 1

Answer: (E)

Concept: The total cell EMF ($E_{cell}$) is influenced by the concentrations of the anode and cathode solutions via the Nernst equation.
• Cell Reaction: $M(s) + X^{2+}(aq) \rightarrow M^{2+}(aq) + X(s)$.
• Formula: $E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \frac{[Anode \text{ ion}]}{[Cathode \text{ ion}]}$.
• Rule: $E_{cell}$ increases when the cathode concentration increases or the anode concentration decreases.

Step 1: Evaluate the concentration ratios ($Q = \frac{[M^{2+}]}{[X^{2+}]}$). - (i) $Q_1 = 0.1 / 0.01 = 10$. - (ii) $Q_2 = 0.1 / 0.1 = 1$. - (iii) $Q_3 = 0.01 / 0.1 = 0.1$.

Step 2: Compare the EMF values. Since $E_{cell} = E^0 - (\text{constant}) \log Q$: - A larger $Q$ results in a smaller $E_{cell}$. - A smaller $Q$ results in a larger $E_{cell}$. Ordering by $Q$: $Q_1 (10) > Q_2 (1) > Q_3 (0.1)$. Therefore, the EMF order is: $E_3 > E_2 > E_1$.