Q19 / 180 · COMEDK 2024 (Morning Shift) · Chemistry
What would be the EMF of the cell in which the following reaction occurs:
$\begin{aligned}
& \mathrm{Cd}(\mathrm{S})+2 \mathrm{H}^{+} \rightarrow \mathrm{Cd}^{2+}+\mathrm{H}_{2(\mathrm{~g})} \\
& {\left[\mathrm{H}^{+}\right]=0.02 \mathrm{M} \quad \mathrm{E}^0\left(\mathrm{Cd}^{2+} / \mathrm{Cd}\right)=-0.4 \mathrm{~V},\left[\mathrm{Cd}^{2+}\right]=0.01 \mathrm{M} \text { and partial pressure of } \mathrm{H}_2 \text { gas }=0.8 \mathrm{~atm} .}
\end{aligned}$
- A0.4859 V
- B0.4471 V
- C0.3020 V
- Correct answerD0.3616 V
Solution
The cell reaction is $\mathrm{Cd}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Cd}^{2+}(aq) + \mathrm{H}_{2}(g)$.
The standard cell potential $E^{0}_{cell}$ is calculated as $E^{0}_{cathode} - E^{0}_{anode}$.
Given $E^{0}(\mathrm{H}^{+}/\mathrm{H}_{2}) = 0.00 \text{ V}$ and $E^{0}(\mathrm{Cd}^{2+}/\mathrm{Cd}) = -0.40 \text{ V}$.
$E^{0}_{cell} = 0.00 - (-0.40) = 0.40 \text{ V}$.
Using the Nernst equation at $298 \text{ K}$:
$E_{cell} = E^{0}_{cell} - \dfrac{0.0591}{n} \log \dfrac{[\mathrm{Cd}^{2+}] P_{\mathrm{H}_{2}}}{[\mathrm{H}^{+}]^{2}}$
Here $n = 2$, $[\mathrm{Cd}^{2+}] = 0.01 \text{ M}$, $[\mathrm{H}^{+}] = 0.02 \text{ M}$, and $P_{\mathrm{H}_{2}} = 0.8 \text{ atm}$.
$E_{cell} = 0.40 - \dfrac{0.0591}{2} \log \dfrac{0.01 \times 0.8}{(0.02)^{2}}$
$E_{cell} = 0.40 - 0.02955 \log \dfrac{0.008}{0.0004}$
$E_{cell} = 0.40 - 0.02955 \log(20)$
Since $\log(20) = \log(2 \times 10) = 0.3010 + 1 = 1.3010$.
$E_{cell} = 0.40 - 0.02955 \times 1.3010$
$E_{cell} = 0.40 - 0.03844 = 0.36156 \text{ V} \approx 0.3616 \text{ V}$.
Answer: 0.3616 V
The standard cell potential $E^{0}_{cell}$ is calculated as $E^{0}_{cathode} - E^{0}_{anode}$.
Given $E^{0}(\mathrm{H}^{+}/\mathrm{H}_{2}) = 0.00 \text{ V}$ and $E^{0}(\mathrm{Cd}^{2+}/\mathrm{Cd}) = -0.40 \text{ V}$.
$E^{0}_{cell} = 0.00 - (-0.40) = 0.40 \text{ V}$.
Using the Nernst equation at $298 \text{ K}$:
$E_{cell} = E^{0}_{cell} - \dfrac{0.0591}{n} \log \dfrac{[\mathrm{Cd}^{2+}] P_{\mathrm{H}_{2}}}{[\mathrm{H}^{+}]^{2}}$
Here $n = 2$, $[\mathrm{Cd}^{2+}] = 0.01 \text{ M}$, $[\mathrm{H}^{+}] = 0.02 \text{ M}$, and $P_{\mathrm{H}_{2}} = 0.8 \text{ atm}$.
$E_{cell} = 0.40 - \dfrac{0.0591}{2} \log \dfrac{0.01 \times 0.8}{(0.02)^{2}}$
$E_{cell} = 0.40 - 0.02955 \log \dfrac{0.008}{0.0004}$
$E_{cell} = 0.40 - 0.02955 \log(20)$
Since $\log(20) = \log(2 \times 10) = 0.3010 + 1 = 1.3010$.
$E_{cell} = 0.40 - 0.02955 \times 1.3010$
$E_{cell} = 0.40 - 0.03844 = 0.36156 \text{ V} \approx 0.3616 \text{ V}$.
Answer: 0.3616 V
Hence, the correct answer is option: 'D'.