The ${E^ \circ }$ of ${\rm{F}}{{\rm{e}}^{2 + }}/{\rm{Fe}}$ and ${\rm{S}}{{\rm{n}}^{2 + }}/{\rm{Sn}}$ are -0.44 V and -0.14 V respectively. If cell reaction is ${\rm{Fe}} + {\rm{S}}{{\rm{n}}^{2 + }} \to {\rm{F}}{{\rm{e}}^{2 + }} + {\rm{Sn}}$ then emf of the cell is

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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The

{E^ \circ }

\mathrm{F}{\mathrm{e}^{2 + }}/\mathrm{Fe}

and

\mathrm{S}{\mathrm{n}^{2 + }}/\mathrm{Sn}

are -0.44 V and -0.14 V respectively. If cell reaction is

\mathrm{Fe} + \mathrm{S}{\mathrm{n}^{2 + }} \to \mathrm{F}{\mathrm{e}^{2 + }} + \mathrm{Sn}

then emf of the cell is

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Correct option: A

+0.30 V

Option A is correct because the cell emf is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode: E°cell = E°cathode - E°anode.

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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