Question

For the cell reaction \( 4Br^- + O_2 + 4H^+ \rightarrow 2Br_2 + 2H_2O \) at \( 298\,K \), the \( E^0_{\text{cell}} = 0.16\,V \). What would be the \( K_c \) (Equilibrium constant) value if the reverse reaction were to take place?

• A. \( 2.012 \times 10^{-10} \)
• B. \( 8.47 \times 10^{-9} \)
• C. \( 1.422 \times 10^{-11} \)
• D. \( 7.031 \times 10^{-10} \)

Hint:

For electrochemical cells, use \( E^0_{\text{cell}}=\frac{0.0591}{n}\log K \) at \(298K\). For reverse reaction, equilibrium constant becomes reciprocal.

Answer 1

The Correct Option is C

Step 1: Write relation between \( E^0_{\text{cell}} \) and equilibrium constant.
At \( 298\,K \):
\[ E^0_{\text{cell}} = \frac{0.0591}{n}\log K \]

Step 2: Identify number of electrons transferred.
For the given reaction:
\[ 4Br^- \rightarrow 2Br_2 + 4e^- \]
So:
\[ n = 4 \]

Step 3: Substitute values for forward reaction.
\[ 0.16 = \frac{0.0591}{4}\log K \]

Step 4: Solve for \( \log K \).
\[ \log K = \frac{0.16 \times 4}{0.0591} \]
\[ \log K \approx 10.83 \]

Step 5: Find \( K \) for forward reaction.
\[ K_{\text{forward}} = 10^{10.83} \]
\[ K_{\text{forward}} \approx 7.03 \times 10^{10} \]

Step 6: Find equilibrium constant for reverse reaction.
For reverse reaction:
\[ K_{\text{reverse}} = \frac{1}{K_{\text{forward}}} \]
\[ K_{\text{reverse}} = \frac{1}{7.03 \times 10^{10}} \]
\[ K_{\text{reverse}} \approx 1.422 \times 10^{-11} \]

Step 7: Final conclusion.
\[ \boxed{1.422 \times 10^{-11}} \]