For the following reactions:
$Z n \to Z n^{2 +} + 2 e^{-}; E = + 0.76 V$
$A u \to A u^{3 +} + 3 e^{-}; E = - 1.42 V$
If gold foil is placed in a solution containing $Z n^{2 +}$ , the reaction potential would be:

- 1.34 V

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- 2.18 V

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- 0.66 V

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+ 2.18 V

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+ 1.34 V

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Solution

The correct option is B $- 2.18 V$
If gold foil is placed in a solution containing $Z n^{2 +}$ , Gold will be reduced and Zinc will be oxidised and the following reaction will take place:
$3 Z n + 2 A u^{3 +} \to 3 Z n^{2 +} + 2 A u$
The reaction potential for the above reaction is given by :
$E = E_{c a t h o d e} - E_{a n o d e} = - 1.42 - 0.76 = - 2.18 V$

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Similar questions

Q. Electrode potentials are:

{Z n}^{0} \to {Z n}^{2 +} + 2 e^{-}; E^{0} = + 0.76 V

{A u}^{0} \to {A u}^{3 +} + 3 e^{-}; E^{0} = - 1.42 V

What is the potential of the reaction, if a gold foil is kept in a solution containing zinc ions?

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:

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

will

Calculate

E^{\circ}

values of the following reactions.

2 A g + Z n^{2 +} \to 2 A g^{+} + Z n

Given that:

E_{A g^{+} / A g}^{\circ} = 0.80 V, E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V

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For the following reactions:Zn→Zn2++2e−;E=+0.76VAu→Au3++3e−;E=−1.42VIf gold foil is placed in a solution containing Zn2+, the reaction potential would be:

For the following reactions:
$Z n \to Z n^{2 +} + 2 e^{-}; E = + 0.76 V$
$A u \to A u^{3 +} + 3 e^{-}; E = - 1.42 V$
If gold foil is placed in a solution containing $Z n^{2 +}$ , the reaction potential would be: