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: Cell reaction and Nernst equation.
Cell reaction: $Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)$. The Nernst equation is: $E_{cell} = E^\circ_{cell} - \dfrac{0.059}{2}\log Q$, where $Q = \dfrac{[Mg^{2+}]}{[Cu^{2+}]}$.
Step 2: How to increase $E_{cell}$.
$E_{cell}$ increases when $Q$ decreases (since we subtract a term containing $\log Q$). $Q$ decreases when $[Mg^{2+}]$ (numerator) decreases or $[Cu^{2+}]$ (denominator) increases.
Step 3: Evaluate each option.
Decreasing $[Mg^{2+}]$ from 1.0 M to 0.1 M reduces $Q$ from 1 to 0.1, so $\log Q$ becomes negative, increasing $E_{cell}$. Decreasing $[Cu^{2+}]$ would increase $Q$ and decrease $E_{cell}$.
Step 4: Conclusion.
Decreasing only $[Mg^{2+}]$ to 0.1 M increases the EMF.
\[ \boxed{\text{Decrease only } [Mg^{2+}] \text{ to 0.1 M}} \]