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  • For the given reactions <img alt="\mathrm{Sn ^{2+}+2 e ^{-} \rightarrow Sn}" src="/latex-image/?%5Cmathrm%7BSn%20%5E%7B2&plus;%7D&plus;2%20e%20%5E%7B-%7D%20%5Crightarrow%20Sn%7D"

For the given reactions

\mathrm{Sn ^{2+}+2 e ^{-} \rightarrow Sn}

\mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \rightarrow \mathrm{Sn}

the electrode potentials are; \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\mathrm{O}}=-0.140 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{Sn}^{4+} / \mathrm{Sn}}^{\mathrm{S}}=0.010 \mathrm{~V}. The magnitude of standared electrode potential for \mathrm{\mathrm{Sn}^{4+} | \mathrm{Sn}^{2+} \text { i.e. } \mathrm{E}_{\mathrm{Sn}^{4+}| \mathrm{Sn}^{2+}}^{\mathrm{0}} \text { is }} __________ \times 10^{-2} \mathrm{~V}.

(Nearest integer)

Option: 1

16


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}, \mathrm{~E_{Sn^{2+}|Sn}^0=-0.14V,~ \Delta G_1^0=-2F(-0.14)}\quad -\text{(i)}\mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \rightarrow \mathrm{Sn}, \mathrm{~E_{Sn^{4+}|Sn}^0=0.01V,~ \Delta G_2^0=-4F(0.01)}\quad -\text{(ii)}

Operation \mathrm{(ii)-(i)} gives us 

\mathrm{Sn^{4+}+2e^-\rightarrow Sn^{2+},E^0_{Sn^{4+}|Sn^{2+}}, \Delta G^0_3=-2FE^0} 

Thus, we can write 

\begin{aligned} &\mathrm{\Delta G_{3}^{0}=\Delta G_{2}^{0}-\Delta G_{1}^{0}} \\ \\&\mathrm{-2 E^{0} \times F=-(0.04+0.28) \times F} \end{aligned}

\mathrm{E^{\circ}=0.16 \text { Volt }=16 \times 10^{-2} \mathrm{~V}}

Hence, 16 is correct answer

Posted by

Kuldeep Maurya

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