For $\mathrm{Zn}^{2+} \mid \mathrm{Zn}, E^{\circ}=-0.76 \mathrm{~V}$ then EMF of the cell $\mathrm{Zn} |…

For $\mathrm{Zn}^{2+} \mid \mathrm{Zn}, E^{\circ}=-0.76 \mathrm{~V}$ then EMF of the cell $\mathrm{Zn} | \mathrm{Zn}^{2+}(1 \mathrm{M})\left|2 \mathrm{H}^{+}(1 \mathrm{M})\right| \mathrm{H}_2$ (1 atm) will be
  1. $-0.76 \mathrm{~V}$
  2. $0.76 \mathrm{~V}$
  3. $0.38 \mathrm{~V}$
  4. $-0.38 \mathrm{~V}$

Solution

\(\begin{aligned} & E_{\text {cell }}^{\circ}=E_{\text {right }}^{\circ}-E_{\text {left }}^{\circ} \\ & E_{\text {cell }}^{\circ}=0-(0-0.76)=0.76 \mathrm{~V}\end{aligned}\)

Asked in: NEET 2016 (Phase 2)

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