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Electrochemistry — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry3 Marks
Question
Calculate the equilibrium constant for the reaction,$\text{Cd}^{2+}(\text{aq})+\text{Zn(s)}\xrightarrow{ \ \ \ \ \ \ \ }\text{Zn}^{2+}(\text{aq) + Cd(s)}$
If $\text{E}^{\circ}_{\text{Cd}^{2+}/\text{Cd}}=-0.403\text{V;}\text{ E}^{\circ}_{\text{Zn}^{2+}/\text{Zn}}=-0.763\text{V}$
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Answer

$\text{E}^{\circ}_{\text{cell}}=\text{E}^{\circ}_{\text{Cd}^{2+}/\text{Cd}}-\text{E}^{\circ}_{\text{Zn}^{2+}/\text{Zn}}$
$=-0.403\text{V}-(-0.763\text{V})=0.360\text{V, n}=2$
$\log\text{K}_{\text{c}}=\Big(\frac{\text{nE}^{\circ}_{\text{cell}}}{0.059}\Big)=\Big(\frac{2\times0.360}{0.059}\Big)=\Big(\frac{0.720}{0.059}\Big)=2=12.20$
$\text{K}_{\text{c}}=\text{antilog}(12.20)=1.585\times10^{12}$

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