The standard oxidation potentials Eo- for the half cell reactions are given asZn - Zn 2- -2e- - E- - - 0-76 VFe - Fe 2 - -2e- E- - - 0-41 VThe EMF for the cell reaction-Fe2- - Zn - Zn2- Fe is-

Oxidation potentials are given for the two half cell reactions: Fe → Fe^ 2 + +2e^ , E^° = + 0.41 V ...........(i) Zn → Zn ^2+ +2e^ , E^° = + 0.76 V....... ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Oxidation and Reduction

The standard ...

Question

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:

- 0.35 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 0.35 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.17 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 1.17 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $+ 0.35 V$
Oxidation potentials are given for the two half cell reactions:
$F e \to F e^{2 +} + 2 e^{-}, E^{\circ} = + 0.41 V . . . . . . . . . . . (i)$
$Z n \to Z n^{2 +} + 2 e^{-}, E^{\circ} = + 0.76 V . . . . . . . . . . . . (i i)$
Subtracting $(i i)$ from $(i)$ , we can have the desired reaction
$\Rightarrow F e^{2 +} + Z n \to Z n^{2 +} + F e, E^{\circ} = (0.76 - 0.41) V = + 0.35 V$

Suggest Corrections

Similar questions

Q. 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)

Q. The standard oxidation potentials,

E^{\circ}

, for the half reactions are as,

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = + 0.76 v o l t

F e \to F e^{2 +} + 2 e^{-}; E^{\circ} = + 0.41 v o l t

The emf of the cell,

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:

Join BYJU'S Learning Program

The standard oxidation potentials Eo, for the half cell reactions are given as Zn→Zn2++2e−, E∘=+0.76 VFe→Fe2++2e−, E∘=+0.41 V The EMF for the cell reaction, Fe2++Zn→Zn2++Fe is:

The correct option is B +0.35 VOxidation potentials are given for the two half cell reactions: Fe→Fe2++2e−, E∘=+0.41 V...........(i) Zn→Zn2++2e−, E∘=+0.76 V............(ii) Subtracting (ii) from (i), we can have the desired reaction ⇒Fe2++Zn→Zn2++Fe, E∘=(0.76−0.41) V =+0.35 V

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: