JEE MAIN - Chemistry (2022 - 28th June Evening Shift - No. 15)
For the given reactions
Sn2+ + 2e$$-$$ $$\to$$ Sn
Sn4+ + 4e$$-$$ $$\to$$ Sn
the electrode potentials are ; $$E_{S{n^{2 + }}/Sn}^o = - 0.140$$ V and $$E_{S{n^{4 + }}/Sn}^o = + 0.010$$ V. The magnitude of standard electrode potential for $$S{n^{4 + }}/S{n^{2 + }}$$ i.e. $$E_{S{n^{4 + }}/S{n^{2 + }}}^o$$ is _____________ $$\times$$ 10$$-$$2 V. (Nearest integer)
Answer
16
Explanation
$\mathrm{Sn} \longrightarrow \mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \quad \mathrm{E}_{1}^{0}=0.140 \mathrm{~V}$
$$ \mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Sn} \quad \mathrm{E}_{2}^{0}=0.010 \mathrm{~V} $$
$$ \begin{aligned} & \mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+} \quad \mathrm{E}_{\text {cell }}^{0} \\\\ & \mathrm{E}_{\mathrm{cell}}^{\mathrm{O}}=\frac{\mathrm{n}_{2} \mathrm{E}_{2}^{\mathrm{o}}+\mathrm{n}_{1} \mathrm{E}_{1}^{0}}{\mathrm{n}}=\frac{4(0.010)+2(0.140)}{2} \\\\ & \mathrm{E}_{\text {cell }}^{0}=0.16 \mathrm{~V}=16 \times 10^{-2} \mathrm{~V} \end{aligned} $$
$$ \mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Sn} \quad \mathrm{E}_{2}^{0}=0.010 \mathrm{~V} $$
$$ \begin{aligned} & \mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+} \quad \mathrm{E}_{\text {cell }}^{0} \\\\ & \mathrm{E}_{\mathrm{cell}}^{\mathrm{O}}=\frac{\mathrm{n}_{2} \mathrm{E}_{2}^{\mathrm{o}}+\mathrm{n}_{1} \mathrm{E}_{1}^{0}}{\mathrm{n}}=\frac{4(0.010)+2(0.140)}{2} \\\\ & \mathrm{E}_{\text {cell }}^{0}=0.16 \mathrm{~V}=16 \times 10^{-2} \mathrm{~V} \end{aligned} $$
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