Reduction potential for the following half-cell reaction are Zn - Zn2-2e- E0Zn2-Zn-0-76V Fe - Fe2-2e- E0-0-41 V The EMF for the cell reaction Fe2-Zn- Zn2-Fe will

The correct option is B +0.35 V Fe^2++Zn→ Zn^2++Fe Zn→ Zn^2++2e^ #8722; ; E^ #8861;=+0.76V Fe→ Fe^2++2e^ #8722; #160;; E^ ...

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Byju's Answer

Standard XII

Chemistry

EMF

Reduction pot...

Question

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

+ 0.35

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Solution

The correct option is B $+ 0.35$ V
$F e^{2 +} + Z n \to Z n^{2 +} + F e$
$Z n \to Z n^{2 +} + 2 e^{-}$
; $E^{⊝} = + 0.76 V$
$F e \to F e^{2 +} + 2 e^{-}$
; $E^{⊝} = + 0.41 V$
These are oxidation potentials.
Reduction potentials are equal and opposite.
Fe forms cathode and Zn forms anode.
$E_{c e l l}^{⊝} = (E_{r e d}^{⊝}) c + (E_{o x i d}^{⊝}) a$
$= (- 0.41 + 0.76) = 0.35 V$

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Similar questions

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. 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 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

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Reduction potential for the following half/cell reaction are Zn→Zn2++2e−(E0(Zn2+/Zn)=−0.76V)Fe→Fe2++2e−E0=0.41 V The EMF for the cell reaction Fe2++Zn→Zn2++Fe will

The correct option is B +0.35 VFe2++Zn→Zn2++FeZn→Zn2++2e−;E⊝=+0.76VFe→Fe2++2e− ;E⊝=+0.41VThese are oxidation potentials.Reduction potentials are equal and opposite. Fe forms cathode and Zn forms anode.E⊝cell=(E⊝red)c+(E⊝oxid)a=(−0.41+0.76)=0.35V

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