The standard reduction potential $E^{o}$ for the half reactions are as
$Z n ⟶ {Z n}^{2 +} + 2 e^{-}$ ; $E^{o} = + 0.76 V$
$F e ⟶ {F e}^{2 +} + 2 e^{-}$ ; $E^{o} = + 0.41 V$
The emf for cell reaction,
${F e}^{2 +} + Z n ⟶ {Z n}^{2 +} + F e$ , is

+ 0.35 V

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- 0.35 V

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- 1.17 V

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+ 1.17 V

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Solution

The correct option is A $+ 0.35 V$
$E M F = E_{c a t h o d e} - E_{a n o d e}$
$= - 0.41 - (- 0.76)$
$E M F = + 0.35 V$

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Similar questions

Q. The standard reduction potential

E^{o}

for half reactions are,

Z n \to Z n^{2 +} + 2 e^{-}; E^{o} = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E^{o} = + 0.41 V

The emf of the cell reaction;

F e^{2 +} + Z n \to Z n^{2 +} + F e

is:

Q. The standard reduction potential

E^{o}

for half reaction are:

Z n \to {Z n}^{2 +} + 2 e^{-}; E^{o} = + 0.76 V

F e \to {F e}^{2 +} + 2 e^{-}; E^{o} = + 0.41 V

The EMF of the cell reaction is:

{F e}^{2 +} + Z n \to {Z n}^{2 +} + F e

The standard reduction potentials, $E^{\circ}$ , for the half-reactions are as $Z n = Z n^{2 +} + 2 e$ , $E^{\circ}$ = + 0.76 V and $F e = F e^{2 +} + 2 e$ , $E^{\circ}$ = + 0.41 V. The e.m.f. for the cell reaction, $F e^{2 +} + Z n = Z n^{2 +} + F e$ is