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

Concept:

• The emf of a cell under non-standard conditions is determined by the Nernst Equation.

• For the cell reaction: \( Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s) \), the number of electrons transferred (\( n \)) is 2.

• The Nernst equation is: \[ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{n} \log \frac{[Mg^{2+}]}{[Cu^{2+}]} \]

• To increase \( E_{\text{cell}} \), we must decrease the value of the term being subtracted.
Step 1: Analyze the logarithmic term
The expression \( \log \frac{[Mg^{2+}]}{[Cu^{2+}]} \) must be made smaller (or more negative) to increase the total cell potential.
This can be achieved by:
1. Decreasing the concentration of products (Anode compartment ions, \( [Mg^{2+}] \)).
2. Increasing the concentration of reactants (Cathode compartment ions, \( [Cu^{2+}] \)).

Step 2: Test Option (A)
Decrease \( [Mg^{2+}] \) from \( 1.0 \, M \) to \( 0.1 \, M \):
\[ E = E^\circ - \frac{0.059}{2} \log \frac{0.1}{1.0} \] \[ E = E^\circ - \frac{0.059}{2} (-1) = E^\circ + 0.0295 \, V \] The emf increases.

Step 3: Test Option (B)
Decrease \( [Cu^{2+}] \) from \( 1.0 \, M \) to \( 0.1 \, M \):
\[ E = E^\circ - \frac{0.059}{2} \log \frac{1.0}{0.1} \] \[ E = E^\circ - \frac{0.059}{2} (1) = E^\circ - 0.0295 \, V \] The emf decreases.

Step 4: Conclusion
Decreasing the concentration of the oxidation product (\( Mg^{2+} \)) reduces the "back-pressure" on the reaction, shifting the equilibrium to the right and increasing the potential.