Question

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

• A. Decrease only $[Mg^{2+}]$ to 0.1 M
• B. Decrease only $[Cu^{2+}]$ to 0.1 M
• C. Increase both $[Mg^{2+}]$ and $[Cu^{2+}]$ to 2.0 M
• D. Increase only $[Mg^{2+}]$ to 2.0 M

Hint:

To increase $E_{cell}$: Decrease $Q$ by reducing product concentration ($[Mg^{2+}]$) or increasing reactant concentration ($[Cu^{2+}]$).

Answer 1

The Correct Option is A

Step 1: Concept
The cell reaction is: $Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)$. By the Nernst equation: $E_{cell} = E^\circ_{cell} - \dfrac{0.059}{n}\log Q$, where $Q = \dfrac{[Mg^{2+}]}{[Cu^{2+}]}$.

Step 2: Analysis
To increase $E_{cell}$, we need to decrease $Q$. $Q$ decreases when $[Mg^{2+}]$ (numerator) decreases or when $[Cu^{2+}]$ (denominator) increases.

Step 3: Conclusion
Decreasing $[Mg^{2+}]$ to 0.1 M lowers the reaction quotient $Q$, which raises the cell emf.

Final Answer: (A)