Question

What is the reduction potential of a half-cell consisting of a Pt electrode dipped in \(2.2M\ Fe^{2+}\) and \(0.04M\ Fe^{3+}\) solution where the reaction taking place is conversion of \(Fe^{3+}\) ions to \(Fe^{2+}\)? Given \(E^\circ(Fe^{3+}/Fe^{2+})=0.771\ V\)

• A. \(0.598\ V\)
• B. \(0.723\ V\)
• C. \(0.668\ V\)
• D. \(0.719\ V\)

Hint:

For \(Fe^{3+}/Fe^{2+}\), always write the reduction reaction first and use \(Q=\frac{[Fe^{2+}]}{[Fe^{3+}]}\) in the Nernst equation.

Answer 1

The Correct Option is C

Step 1: Write the reduction reaction.
The given half-cell reaction is:
\[ Fe^{3+}+e^- \rightarrow Fe^{2+} \]
Here, the number of electrons transferred is:
\[ n=1 \]

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

Step 3: Substitute the given concentrations.
\[ [Fe^{2+}] = 2.2M,\quad [Fe^{3+}] = 0.04M \]
So,
\[ Q=\frac{2.2}{0.04} \]
\[ Q=55 \]

Step 4: Substitute values in Nernst equation.
\[ E = 0.771 - \frac{0.0591}{1}\log(55) \]
\[ E = 0.771 - 0.0591\log(55) \]

Step 5: Evaluate logarithmic value.
\[ \log(55) \approx 1.740 \]
Therefore,
\[ 0.0591 \times 1.740 \approx 0.103 \]

Step 6: Calculate reduction potential.
\[ E = 0.771 - 0.103 \]
\[ E = 0.668\ V \]

Step 7: Final conclusion.
Thus, the reduction potential of the half-cell is:
\[ \boxed{0.668\ V} \]