The emf of the cell ${\rm{Ni}}\left| {{\rm{N}}{{\rm{i}}^{2 + }}\left( {1.0{\rm{M}}} \right)} \right|\left| {{\rm{A}}{{\rm{u}}^{3 + }}\left( {1.0{\rm{M}}} \right)} \right|{\rm{Au}}$ is [${{\rm{E}}^ \circ }\left( {{\rm{N}}{{\rm{i}}^{2 + }}/{\rm{Ni}}} \right) = $ -0.25 V and ${E^ \circ }\left( {{\rm{A}}{{\rm{u}}^{3 + }}/{\rm{Au}}} \right) = $ +1.5 V]

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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The emf of the cell

\mathrm{Ni}\left| {\mathrm{N}{\mathrm{i}^{2 + }}\left( {1.0\mathrm{M}} \right)} \right|\left| {\mathrm{A}{\mathrm{u}^{3 + }}\left( {1.0\mathrm{M}} \right)} \right|\mathrm{Au}

is [

{\mathrm{E}^ \circ }\left( {\mathrm{N}{\mathrm{i}^{2 + }}/\mathrm{Ni}} \right) =

-0.25 V and

{E^ \circ }\left( {\mathrm{A}{\mathrm{u}^{3 + }}/\mathrm{Au}} \right) =

+1.5 V]

View answer and explanation

Correct option: D

1.75 V

Option D (1.75 V) is correct because it is the difference between the reduction potentials of the cathode and anode.