The standard oxidation potentials- Eo- for the half cell reactions are as follows-Zn - Zn2-2e- Eo-0-76 VFe - Fe2-2e- Eo-0-41 VThe EMF for the cell oxidation-Fe2- - Zn - Zn2- - FeGive your answer upto two decimal only

We know that, Electrode with higher oxidation potential acts as an anode E^o_cell= nbsp;(E^o_red(cathode) E^o_red(anode) =0.41 ( 0.76)=0.41+0.76=0.35 V ...

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Standard IX

Chemistry

Electrode Potential

The standard ...

Question

The standard oxidation potentials, $E^{o}$ , for the half cell reactions are as follows:
$Z n \to Z n^{2 +} + 2 e^{-}, E^{o} = 0.76 V F e \to F e^{2 +} + 2 e^{-}, E^{o} = 0.41 V$
The EMF for the cell oxidation,
$F e^{2 +} + Z n \to Z n^{2 +} + F e$ (Give your answer upto two decimal only)

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Solution

We know that,
Electrode with higher oxidation potential acts as an anode
$E_{c e l l}^{o} = (E_{r e d (c a t h o d e)}^{o} - E_{r e d (a n o d e)}^{o} = 0.41 - (- 0.76) = 0.41 + 0.76 = 0.35 V$

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Q. The standard oxidation potentials

E^{o}

, for the half cell reactions are given as

Z n \to Z n^{2 +} + 2 e^{-}, E^{\circ} = + 0.76 V F e \to F e^{2 +} + 2 e^{-}, E^{\circ} = + 0.41 V

The EMF for the cell reaction,

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

is:

Q. Given the standard half-cell potentials

(E^{o})

of the following as

Z n = Z n^{2 +} + 2 e^{-}

;

E^{o} = + 0.76

F e = F e^{2 +} + 2 e^{-}

;

E^{o} = + 0.41

Then the standard e.m.f. of the cell with the reaction

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

is?

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

, is

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

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The standard oxidation potentials, Eo, for the half cell reactions are as follows: Zn→Zn2++2e−, Eo=0.76 VFe→Fe2++2e−, Eo=0.41 V The EMF for the cell oxidation, Fe2++Zn→Zn2++Fe(Give your answer upto two decimal only)

We know that, Electrode with higher oxidation potential acts as an anode Eocell=(Eored(cathode)−Eored(anode)=0.41−(−0.76)=0.41+0.76=0.35 V

The standard oxidation potentials, $E^{o}$ , for the half cell reactions are as follows:
$Z n \to Z n^{2 +} + 2 e^{-}, E^{o} = 0.76 V F e \to F e^{2 +} + 2 e^{-}, E^{o} = 0.41 V$
The EMF for the cell oxidation,
$F e^{2 +} + Z n \to Z n^{2 +} + F e$ (Give your answer upto two decimal only)

We know that,
Electrode with higher oxidation potential acts as an anode
$E_{c e l l}^{o} = (E_{r e d (c a t h o d e)}^{o} - E_{r e d (a n o d e)}^{o} = 0.41 - (- 0.76) = 0.41 + 0.76 = 0.35 V$