Question

Consider the following cell at \( 298 \, K \): \( Mg (s) \mid Mg^{2+} (1.0 \, M) \parallel Cu^{2+} (1.0 \, M) \mid Cu (s) \). How can we increase the emf of the cell using the same substances ?

• A. \( \text{By decreasing only the } [Mg^{2+}] \text{ to } 0.1 \, M \)
• B. \( \text{By decreasing only the } [Cu^{2+}] \text{ to } 0.1 \, M \)
• C. \( \text{By increasing both } [Mg^{2+}] \text{ and } [Cu^{2+}] \text{ to } 2.0 \, M \)
• D. \( \text{By increasing only the } [Mg^{2+}] \text{ to } 2.0 \, M \)

Hint:

Emf increases if:
1. Reactant (Cathode ion) concentration increases.
2. Product (Anode ion) concentration decreases.
Think of it as pushing the reaction forward!

Answer 1

The Correct Option is A

Step 1: Write the Nernst equation for this cell.
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{2}\log\frac{[Mg^{2+}]}{[Cu^{2+}]} \] since two electrons are exchanged in $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$.
Step 2: See what needs to happen to the log term to raise $E_{cell}$.
Because that log term is being subtracted, $E_{cell}$ goes up only when $\log\frac{[Mg^{2+}]}{[Cu^{2+}]}$ becomes smaller, which means the ratio $[Mg^{2+}]/[Cu^{2+}]$ itself must shrink.
Step 3: Check which change achieves that.
Dropping $[Mg^{2+}]$ to $0.1\,M$ while keeping $[Cu^{2+}]$ at $1.0\,M$ makes the ratio ten times smaller, giving \[ E = E^\circ - \frac{0.0591}{2}\log(0.1) = E^\circ + 0.0295\,V \] which is a genuine increase. Dropping $[Cu^{2+}]$ instead would make the ratio bigger and actually lower $E_{cell}$.
Step 4: State the answer.
Reducing the anode's own product concentration pulls the reaction further to the right and raises the cell's voltage. \[ \boxed{\text{Decrease } [Mg^{2+}] \text{ to } 0.1\,M} \]