Q9 / 90 · JEE Main 2019 (11 Jan Shift 1) · Chemistry
For the cell $\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(\mathrm{aq}) \| \mathrm{M}^{\mathrm{x}+}(\mathrm{aq})\right| \mathrm{M}(\mathrm{s}),$ different half
cells and their standard electrode potentials are given below:
If $\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V},$ which cathode will give a maximum value of $E_{\text {cell }}^{0}$ per electron transferred?
If $\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V},$ which cathode will give a maximum value of $E_{\text {cell }}^{0}$ per electron transferred?
- Correct answerA$\mathrm{Ag}^{+} / \mathrm{Ag}$
- B$\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$
- C$\mathrm{Au}^{3+} / \mathrm{Au}$
- D$\mathrm{Fe}^{2+} / \mathrm{Fe}$
Solution
$\mathrm{Zn}+2 \mathrm{Ag}^{+} \longrightarrow 2 \mathrm{Ag}+\mathrm{Zn}^{2+}\left(n=2 e^{-}\right)$
$=0.80-(-0.76)=1.56 \mathrm{~V}$ for $2 e$
$\therefore \mathrm{E}_{\text {cell }}^{\circ}$ for $1 e^{-}=\frac{1.56}{2}=0.78 \mathrm{~V}$
Hence, the correct answer is option: 'A'.