If ${\rm{M}}{{\rm{g}}^{2 + }} + 2e \to {\rm{Mg}}\left( s \right);E = - 2.37{\rm{V}}$ ${\rm{C}}{{\rm{u}}^{2 + }} + 2e \to {\rm{Cu}}\left( s \right);E = + 0.34{\rm{V}}$? The e.m.f. of the cell ${\rm{Mg}}\left| {{\rm{M}}{{\rm{g}}^{2 + }}} \right|\left| {{\rm{C}}{{\rm{u}}^{2 + }}} \right|{\rm{Cu}}$ is:

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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\mathrm{M}{\mathrm{g}^{2 + }} + 2e \to \mathrm{Mg}\left( s \right);E = - 2.37\mathrm{V}

\mathrm{C}{\mathrm{u}^{2 + }} + 2e \to \mathrm{Cu}\left( s \right);E = + 0.34\mathrm{V}

?

The e.m.f. of the cell

\mathrm{Mg}\left| {\mathrm{M}{\mathrm{g}^{2 + }}} \right|\left| {\mathrm{C}{\mathrm{u}^{2 + }}} \right|\mathrm{Cu}

is:

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Correct option: A

2.71 V

Option A (2.71 V) is correct because the EMF of the cell is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode.

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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