Question

For a cell \( 2M(s) + O_2(g) + 4H^+ \rightarrow 2M^{2+}(aq) + 2H_2O(l) \), the \( E^\circ_{\text{cell}} = +1.67\,V \). When \( [M^{2+}] = 1.0 \times 10^{-3} \, M \) and \( p(O_2) = 0.1 \, atm \), the EMF of the cell becomes \( +1.57 \, V \). Calculate the pH of the electrochemical cell.

• A. 3.49
• B. 2.95
• C. 5.24
• D. 12.01

Hint:

In electrochemical pH problems, write \( Q \) carefully and use \( pH=-\log[H^+] \) after applying the Nernst equation.

Answer 1

The Correct Option is B

Step 1: Write Nernst equation.
\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n}\log Q \]

Step 2: Find number of electrons.
For the reaction:
\[ 2M(s) + O_2(g) + 4H^+ \rightarrow 2M^{2+}(aq) + 2H_2O(l) \]
Number of electrons transferred is:
\[ n = 4 \]

Step 3: Write reaction quotient.
\[ Q = \frac{[M^{2+}]^2}{p(O_2)[H^+]^4} \]

Step 4: Substitute in Nernst equation.
\[ 1.57 = 1.67 - \frac{0.0591}{4}\log Q \]
\[ 0.10 = \frac{0.0591}{4}\log Q \]

Step 5: Calculate \( \log Q \).
\[ \log Q = \frac{0.10 \times 4}{0.0591} \]
\[ \log Q \approx 6.77 \]

Step 6: Substitute concentration values.
\[ Q = \frac{(10^{-3})^2}{0.1[H^+]^4} \]
\[ Q = \frac{10^{-6}}{10^{-1}[H^+]^4} \]
\[ Q = \frac{10^{-5}}{[H^+]^4} \]
Taking log:
\[ \log Q = -5 - 4\log[H^+] \]
Since:
\[ pH = -\log[H^+] \]
\[ \log Q = -5 + 4pH \]

Step 7: Final calculation.
\[ 6.77 = -5 + 4pH \]
\[ 4pH = 11.77 \]
\[ pH = 2.94 \approx 2.95 \]
\[ \boxed{2.95} \]