Solve the following


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 $\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\mathrm{O}}=-0.76 \mathrm{~V}$, which cathode will give a maximum value of $E_{\text {cell }}^{\circ}$ per electron transferred?

  1. $\mathrm{Ag}^{+} / \mathrm{Ag}$

  2. $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$

  3. $\mathrm{Au}^{3+} / \mathrm{Au}$

  4. $\mathrm{Fe}^{2+} / \mathrm{Fe}$

Correct Option: 1


$\mathrm{Zn}+2 \mathrm{Ag}^{+} \longrightarrow 2 \mathrm{Ag}+\mathrm{Zn}^{2+} \quad\left(n=2 e^{-}\right)$

$\mathrm{E}_{\text {cell }}^{\circ}=\left(\mathrm{E}_{\text {R.P. }}^{\circ}\right)_{\text {cathode }}-\left(\mathrm{E}_{\text {R.P. }}^{\circ}\right)_{\text {anode }}$

$=0.80-(-0.76)=1.56 \mathrm{~V}$ for $2 e^{-}$

$\therefore \mathrm{E}_{\text {cell }}^{\circ}$ for $1 e^{-}=\frac{1.56}{2}=0.78 \mathrm{~V}$


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