"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 $\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right)$"

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"$\mathrm{A}: \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 e^{-}$

$\mathrm{C}: \mathrm{Sn}^{+\mathrm{x}}+\mathrm{xe}^{-} \rightarrow \mathrm{Sn}$

$$\mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{E}_{\mathrm{Zn} \mid \mathrm{Zn}^{2+}}^{\circ}+\mathrm{E}_{\mathrm{Sn}^{+x} \mid \mathrm{Sn}}^{\circ}$$

$\Rightarrow 0.763+0.008=0.771 \mathrm{~V}$

From the Nernst equation,

$$\mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cell }}^{\circ} \frac{-2.303 \,\mathrm{RT}}{\mathrm{nF}} \log \mathrm{Q}$$

$0.801=0.771-\frac{0.06}{\mathrm{n}} \log 10^{-2}$

$0.03=\frac{0.06}{n} \times 2$

$\mathrm{n}=4$"

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