Question

The EMF of the cell \(Al/Al^{3+}(0.01M) \parallel Fe^{2}+(0.02M)/Fe\) is 1.209 V. The EMF of the cell can be increased by

• A. increasing the concentration of \(Al^{3+}\) and \(Fe^{2+}\)
• B. increasing the concentration of \(Al^{3+}\)
• C. increasing the concentration of \(Fe^{2+}\)
• D. decreasing the concentration of \(Al^{3+}\) and \(Fe^{2+}\)

Hint:

EMF increases when reaction quotient \(Q\) decreases. Increase concentration of reactants or decrease concentration of products.

Answer 1

The Correct Option is C

Step 1: Write the cell reaction.
\[ Al \rightarrow Al^{3+} + 3e^- \quad (\text{anode}) \] \[ Fe^{2+} + 2e^- \rightarrow Fe \quad (\text{cathode}) \]
Overall reaction (after balancing):
\[ 2Al + 3Fe^{2+} \rightarrow 2Al^{3+} + 3Fe \]

Step 2: Write the Nernst equation.
\[ E = E^\circ - \frac{0.0591}{n}\log Q \]
where reaction quotient:
\[ Q = \frac{[Al^{3+}]^2}{[Fe^{2+}]^3} \]

Step 3: Understand effect on EMF.
\[ E \propto -\log Q \]
So, EMF increases when \(Q\) decreases.

Step 4: Analyse effect of concentrations.
\[ Q = \frac{[Al^{3+}]^2}{[Fe^{2+}]^3} \]
To decrease \(Q\):
- Decrease numerator \([Al^{3+}]\) OR
- Increase denominator \([Fe^{2+}]\)

Step 5: Check given options.
(A) Increasing both → effect uncertain
(B) Increasing \([Al^{3+}]\) → increases \(Q\), decreases EMF
(C) Increasing \([Fe^{2+}]\) → decreases \(Q\), increases EMF
(D) Decreasing both → not clearly reducing \(Q\)

Step 6: Correct choice.
Only option (C) clearly decreases \(Q\) and increases EMF.

Step 7: Final conclusion.
\[ \boxed{\text{Increasing the concentration of } Fe^{2+}} \]