The standard reduction potential Eo for half reactions are- Zn- Zn2-2e- Eo-0-76 V Fe2-2e- Fe- Eo-0-41 VThe emf of the cell reaction- Fe2-Zn- Zn2-Fe is-

The correct option is A +1.17 V E_cell=E^o_(cathode) E^o_(anode) E_cell=0.41 ( 0.76) = 1.17 V . ...

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Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

The standard ...

Question

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:

- 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 $+ 1.17 V$
$E_{c e l l} = E_{(c a t h o d e)}^{o} - E_{(a n o d e)}^{o}$

$E_{c e l l} = 0.41 - (- 0.76) = 1.17 V$ .

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Similar questions

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 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

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

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

will

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The standard reduction potential Eo for half reactions are,Zn→Zn2++2e−;Eo=−0.76 VFe2++2e−→Fe;Eo=+0.41 VThe emf of the cell reaction; Fe2++Zn→Zn2++Fe is:

The correct option is A +1.17 VEcell=Eo(cathode)−Eo(anode)Ecell=0.41−(−0.76)=1.17 V.

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:

The correct option is A $+ 1.17 V$
$E_{c e l l} = E_{(c a t h o d e)}^{o} - E_{(a n o d e)}^{o}$

$E_{c e l l} = 0.41 - (- 0.76) = 1.17 V$ .