Electrochemistry 2 Question 7
7. For the cell, $\mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}(a q) | M^{\mathrm{x}+}(a q)\right| M(s)$, different half cells and their standard electrode potentials are given below.
| $\boldsymbol{M}^{\boldsymbol{x}+}(\boldsymbol{a q}) / \boldsymbol{M}(\boldsymbol{s})$ | $\mathrm{Au}^{3+}(a q) /$ $\mathrm{Au}(s)$ |
$\mathrm{Ag}^{+}(a q) /$ $\mathrm{Ag}(s)$ |
$\mathrm{Fe}^{3+}(a q) /$ $\mathrm{Fe}^{2+}(a q)$ |
$\mathrm{Fe}^{2+}(a q) /$ $\mathrm{Fe}(s)$ |
|---|---|---|---|---|
| $\boldsymbol{E}^{\circ} \boldsymbol{M}^{\boldsymbol{x}+} / \boldsymbol{M}$ |
If $E^{\circ}{ }_{\mathrm{Zn}^{2+} / \mathrm{Zn}}=-0.76 \mathrm{~V}$, which cathode will give a maximum value of $E^{\circ}$ cell per electron transferred? (2019 Main, 11 Jan I)
(a) $\frac{\mathrm{Ag}^{+}}{\mathrm{Ag}}$
(b) $\frac{\mathrm{Fe}^{2+}}{\mathrm{Fe}}$
(c) $\frac{\mathrm{Au}^{3+}}{\mathrm{Au}}$
(d) $\frac{\mathrm{Fe}^{3+}}{\mathrm{Fe}^{2+}}$
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Solution:
| Cell Anode (A) Cathode (C) |
$\boldsymbol{E}^{\circ}{ }{\text {cell }}(\mathbf{S R P})=\boldsymbol{E}^{\circ}{ }{\boldsymbol{C}}-\boldsymbol{E}^{\circ}{ }_{\boldsymbol{A}}$ | $\boldsymbol{E}_{\text {cell }}{ }^{\text {free } \overline{\boldsymbol{e}} \text { transfer }}$ |
|---|---|---|
| $\left.\mathrm{Zn} \mathrm{Ag}^{+} \mathrm{Zn}^{+} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{Ag}\right]$ | $0.80-(-0.76)=+1.56 \mathrm{~V}$ for $2 e^{-}$ | $+\frac{1.56}{2}=+0.78 \mathrm{~V}$ |
| $\mathrm{Zn}$ 2. $\left[\mathrm{Zn}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}\right]$ |
$-0.44-(-0.76)=+0.32 \mathrm{~V}$ for $2 e^{-}$ | $+\frac{0.32}{2}=+0.16 \mathrm{~V}$ |
| $\mathrm{Zn}$ 3. $\left[3 \mathrm{Zn}+2 \mathrm{Au}^{3+} \rightarrow 3 n^{2+}+2 \mathrm{Au}\right]$ |
$1.40-(-0.76)=+2.4 \mathrm{~V}$ for $6 e^{-}$ | $+\frac{2.16}{6}=+0.36 \mathrm{~V}$ |
| $\mathrm{Fn}$ 4. $\left[3 \mathrm{Zn}+2 \mathrm{Fe}^{3+} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{Fe}^{2+}\right]$ |
$0.77-(-0.76)=+1.53 \mathrm{~V}$ for $2 e^{-}$ | $+\frac{1.53}{2}=+0.765 \mathrm{~V}$ |
