The standard electrode potential for the half – cell reactions are ${\rm{Z}}{{\rm{n}}^{2 + }} + 2{e^ - } \to {\rm{Zn}};{E^ \circ }$ = -0.76 V ${\rm{F}}{{\rm{e}}^{2 + }} + 2{e^ - } \to {\rm{Fe}};{E^ \circ }$ = -0.44 V The emf of the cell reaction, ${\rm{F}}{{\rm{e}}^{2 + }} + {\rm{Zn}} \to {\rm{Z}}{{\rm{n}}^{2 + }} + {\rm{Fe}}$ is

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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The standard electrode potential for the half – cell reactions are

\mathrm{Z}{\mathrm{n}^{2 + }} + 2{e^ - } \to \mathrm{Zn};{E^ \circ }

= -0.76 V

\mathrm{F}{\mathrm{e}^{2 + }} + 2{e^ - } \to \mathrm{Fe};{E^ \circ }

= -0.44 V
The emf of the cell reaction,

\mathrm{F}{\mathrm{e}^{2 + }} + \mathrm{Zn} \to \mathrm{Z}{\mathrm{n}^{2 + }} + \mathrm{Fe}

View answer and explanation

Correct option: D

+0.32 V

Option D (+0.32 V) is correct because it is the calculated EMF of the cell using the standard reduction potentials.