Q23 / 180 · COMEDK 2023 (Evening Shift) · Chemistry
The reaction taking place in a galvanic cell is as given
$\mathrm{A}(\mathrm{s})+\mathrm{B}^{2+}\left(\mathrm{1} \mathrm{1} \mathrm{1 0} \mathrm{0}^{-\mathrm{M}} \mathrm{M}\right) \rightarrow \mathrm{B}_{(\mathrm{s})}+\mathrm{A}^{2+}(0.1 \mathrm{M}).$
The emf of the cell is $+2.651 \mathrm{~V}$. If the standard emf of the cell is $+2.71 \mathrm{~V}$, what is the value of $\mathrm{X}$ ?
- AX = 2
- Correct answerBX = 3
- CX = 4
- DX = 6
Solution
The cell reaction is $\mathrm{A}(\mathrm{s}) + \mathrm{B}^{2+}(10^{-X} \mathrm{M}) \rightarrow \mathrm{B}(\mathrm{s}) + \mathrm{A}^{2+}(0.1 \mathrm{M})$.
The Nernst equation for the cell at $298 \mathrm{K}$ is given by $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \dfrac{0.0591}{n} \log Q$, where $n$ is the number of electrons transferred.
Here, $n = 2$, $E_{\text{cell}} = 2.651 \mathrm{~V}$, and $E^{\circ}_{\text{cell}} = 2.71 \mathrm{~V}$.
The reaction quotient $Q$ is $\dfrac{[\mathrm{A}^{2+}]}{[\mathrm{B}^{2+}]} = \dfrac{0.1}{10^{-X}} = 10^{X-1}$.
Substituting the values into the Nernst equation:
$2.651 = 2.71 - \dfrac{0.0591}{2} \log(10^{X-1})$
$2.651 - 2.71 = -0.02955 \times (X-1)$
$-0.059 = -0.02955 \times (X-1)$
$X-1 = \dfrac{0.059}{0.02955} \approx 2$
$X = 2 + 1 = 3$.
Answer: X = 3
The Nernst equation for the cell at $298 \mathrm{K}$ is given by $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \dfrac{0.0591}{n} \log Q$, where $n$ is the number of electrons transferred.
Here, $n = 2$, $E_{\text{cell}} = 2.651 \mathrm{~V}$, and $E^{\circ}_{\text{cell}} = 2.71 \mathrm{~V}$.
The reaction quotient $Q$ is $\dfrac{[\mathrm{A}^{2+}]}{[\mathrm{B}^{2+}]} = \dfrac{0.1}{10^{-X}} = 10^{X-1}$.
Substituting the values into the Nernst equation:
$2.651 = 2.71 - \dfrac{0.0591}{2} \log(10^{X-1})$
$2.651 - 2.71 = -0.02955 \times (X-1)$
$-0.059 = -0.02955 \times (X-1)$
$X-1 = \dfrac{0.059}{0.02955} \approx 2$
$X = 2 + 1 = 3$.
Answer: X = 3
Hence, the correct answer is option: 'B'.