JEE MAIN - Chemistry (2022 - 25th July Morning Shift - No. 14)
The cell potential for $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right|\left|\mathrm{Sn}^{x+}\right| \mathrm{Sn}$$ is $$0.801 \mathrm{~V}$$ at $$298 \mathrm{~K}$$. The reaction quotient for the above reaction is $$10^{-2}$$. The number of electrons involved in the given electrochemical cell reaction is ____________.
$$\left(\right.$$ Given $$: \mathrm{E}_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}^{\mathrm{o}}=-0.763 \mathrm{~V}, \mathrm{E}_{\mathrm{Sn}^{x+} \mid \mathrm{Sn}}^{\mathrm{o}}=+0.008 \mathrm{~V}$$ and $$\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right)$$
Answer
4
Explanation
$$\mathrm{A}: \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 e^{-}$$
$$\mathrm{C}: \mathrm{Sn}^{+\mathrm{x}}+\mathrm{xe}^{-} \rightarrow \mathrm{Sn}$$
$$\mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{E}_{\mathrm{Zn} \mid \mathrm{Zn}^{2+}}^{\circ}+\mathrm{E}_{\mathrm{Sn}^{+x} \mid \mathrm{Sn}}^{\circ}$$
$$\Rightarrow 0.763+0.008=0.771 \mathrm{~V}$$
From the Nernst equation,
$$\mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cell }}^{\circ} \frac{-2.303 \,\mathrm{RT}}{\mathrm{nF}} \log \mathrm{Q}$$
$$0.801=0.771-\frac{0.06}{\mathrm{n}} \log 10^{-2}$$
$$0.03=\frac{0.06}{n} \times 2$$
$$\mathrm{n}=4$$
$$\mathrm{C}: \mathrm{Sn}^{+\mathrm{x}}+\mathrm{xe}^{-} \rightarrow \mathrm{Sn}$$
$$\mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{E}_{\mathrm{Zn} \mid \mathrm{Zn}^{2+}}^{\circ}+\mathrm{E}_{\mathrm{Sn}^{+x} \mid \mathrm{Sn}}^{\circ}$$
$$\Rightarrow 0.763+0.008=0.771 \mathrm{~V}$$
From the Nernst equation,
$$\mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cell }}^{\circ} \frac{-2.303 \,\mathrm{RT}}{\mathrm{nF}} \log \mathrm{Q}$$
$$0.801=0.771-\frac{0.06}{\mathrm{n}} \log 10^{-2}$$
$$0.03=\frac{0.06}{n} \times 2$$
$$\mathrm{n}=4$$
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