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EMF

The emf of a ...

Question

The emf of a galvanic cell constituted with the electrodes ${Z n}^{2 +} | Z n (- 0.76 V)$ and ${F e}^{2 +} | F e (- 0.41 V)$ is :

A

- 0.35 V

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B

+ 1.17 V

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C

+ 0.35 V

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D

- 1.17 V

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Solution

The correct option is C $+ 0.35 V$
Emf of the given cell be calculated as:

$E_{c e l l}^{o} = E_{r e d} - E_{o x} = - 0.41 - (- 0.76) = 0.35 V$

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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

Q. Reduction potential for the following half/cell reaction are

Z n \to Z n^{2 +} + 2 e^{-} (E_{(Z n^{2 +} / Z n)}^{0} = - 0.76 V)

F e \to F e^{2 +} + 2 e^{-} E^{0} = 0.41 V

The EMF for the cell reaction

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

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

Q. 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

Q. The standard reduction potentials

E^{⊝}

for the half reactions are follows :

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

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

The EMF for the cell reaction

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

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