Q15 / 180 · COMEDK 2025 (10 May Shift 3) · Chemistry
For a cell $2 \mathrm{M}_{(\mathrm{S})}+\mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$ the $\mathrm{E}^0$ cell $=+1.67 \mathrm{~V}$. When $\left[\mathrm{M}^{2+}\right]$ is $1.0 \times 10^{-3} \mathrm{M}$ and $\mathrm{p}\left(\mathrm{O}_2\right)$ is 0.1 atm , the EMF of the cell becomes +1.57 V . Calculate the pH of the electrochemical cell.
- A$3.49$
- B$12.01$
- C$5.24$
- Correct answerD$2.94$
Solution
The cell reaction is $2M_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2M^{2+}_{(aq)} + 2H_2O_{(l)}$.
The Nernst equation for the cell is $E_{cell} = E^0_{cell} - \dfrac{0.0591}{n} \log Q$, where $n=4$ electrons are transferred.
The reaction quotient $Q$ is given by $Q = \dfrac{[M^{2+}]^2}{p(O_2) [H^+]^4}$.
Substituting the given values: $1.57 = 1.67 - \dfrac{0.0591}{4} \log \left( \dfrac{(10^{-3})^2}{0.1 \times [H^+]^4} \right)$.
$-0.10 = -\dfrac{0.0591}{4} \log \left( \dfrac{10^{-6}}{0.1 \times [H^+]^4} \right)$.
$\dfrac{0.10 \times 4}{0.0591} = \log \left( \dfrac{10^{-5}}{[H^+]^4} \right)$.
$6.768 = \log(10^{-5}) - \log([H^+]^4) = -5 - 4 \log[H^+]$.
Since $pH = -\log[H^+]$, we have $6.768 = -5 + 4(pH)$.
$4(pH) = 11.768$.
$pH = \dfrac{11.768}{4} = 2.942$.
Rounding to the nearest option, the pH is approximately $2.94$.
Answer: $2.94$
The Nernst equation for the cell is $E_{cell} = E^0_{cell} - \dfrac{0.0591}{n} \log Q$, where $n=4$ electrons are transferred.
The reaction quotient $Q$ is given by $Q = \dfrac{[M^{2+}]^2}{p(O_2) [H^+]^4}$.
Substituting the given values: $1.57 = 1.67 - \dfrac{0.0591}{4} \log \left( \dfrac{(10^{-3})^2}{0.1 \times [H^+]^4} \right)$.
$-0.10 = -\dfrac{0.0591}{4} \log \left( \dfrac{10^{-6}}{0.1 \times [H^+]^4} \right)$.
$\dfrac{0.10 \times 4}{0.0591} = \log \left( \dfrac{10^{-5}}{[H^+]^4} \right)$.
$6.768 = \log(10^{-5}) - \log([H^+]^4) = -5 - 4 \log[H^+]$.
Since $pH = -\log[H^+]$, we have $6.768 = -5 + 4(pH)$.
$4(pH) = 11.768$.
$pH = \dfrac{11.768}{4} = 2.942$.
Rounding to the nearest option, the pH is approximately $2.94$.
Answer: $2.94$
Hence, the correct answer is option: 'D'.