The emf of the cell, $\left( {{E_{{\rm{Z}}{{\rm{n}}^{2 + }}/{\rm{Zn}}}} = - 0.76V} \right)$ ${\rm{Zn}}/{\rm{Z}}{{\rm{n}}^{2 + }}\left( {1{\rm{M}}} \right)|\left| {{\rm{C}}{{\rm{u}}^{2 + }}\left( {1{\rm{M}}} \right)} \right|{\rm{Cu}}$ $\left( {{E_{{\rm{C}}{{\rm{u}}^{2 + }}/{\rm{Cu}}}} = + 0.34V} \right)$ will be

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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The emf of the cell,

\left( {{E_{\mathrm{Z}{\mathrm{n}^{2 + }}/\mathrm{Zn}}} = - 0.76V} \right)

\mathrm{Zn}/\mathrm{Z}{\mathrm{n}^{2 + }}\left( {1\mathrm{M}} \right)|\left| {\mathrm{C}{\mathrm{u}^{2 + }}\left( {1\mathrm{M}} \right)} \right|\mathrm{Cu}

\left( {{E_{\mathrm{C}{\mathrm{u}^{2 + }}/\mathrm{Cu}}} = + 0.34V} \right)

will be

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Correct option: A

+1.10 V

Option A is correct. The EMF of the cell is calculated by summing the standard oxidation potential of the zinc electrode and the standard reduction potential of the copper electrode.

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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