${\rm{Ni}}/{\rm{N}}{{\rm{i}}^{2 + }}\left[ {1.0{\rm{M}}} \right]||{\rm{A}}{{\rm{u}}^{3 + }}\left[ {1.0{\rm{M}}} \right]/{\rm{Au}}$ where ${E^ \circ }$ ${\rm{for}}{\kern 1pt} {\rm{N}}{{\rm{i}}^{2 + }}/{\rm{Ni {\kern 1pt} {\kern 1pt} is}} - 0.250{\rm{V}}$ ; and ${E^ \circ }$ for ${\rm{A}}{{\rm{u}}^{3 + }}/{\rm{Au}}$ is 0.150 V. The emf of the cell is

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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\mathrm{Ni}/\mathrm{N}{\mathrm{i}^{2 + }}\left[ {1.0\mathrm{M}} \right]||\mathrm{A}{\mathrm{u}^{3 + }}\left[ {1.0\mathrm{M}} \right]/\mathrm{Au}

where

{E^ \circ }

\mathrm{for}{\kern 1pt} \mathrm{N}{\mathrm{i}^{2 + }}/{\rm{Ni {\kern 1pt} {\kern 1pt} is}} - 0.250\mathrm{V}

; and

{E^ \circ }

for

\mathrm{A}{\mathrm{u}^{3 + }}/\mathrm{Au}

is 0.150 V. The emf of the cell is

View answer and explanation

Correct option: D

+0.4 V

Option D (+0.4 V) is correct because it represents the standard EMF of the cell calculated from the difference in standard reduction potentials.