1755

You visited us 1755 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Chemistry

EMF

Electrode pot...

Question

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?

- 1.34

V (no reaction)

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

+ 1.34

V (no reaction)

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

- 2.18

V (no reaction)

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

+ 2.18

V (reaction)

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

- 0.66

V (reaction)

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

Open in App

Solution

The correct option is C $- 2.18$ V (no reaction)
$E_{c e l l}^{o} = E_{A u}^{o} - E_{Z n}^{o} = - 1.42 - 0.76 = - 2.18 V$
Since, the cell potential is negative, the reaction is non spontaneous. Hence, no reaction will occur.
$A u^{3 +}$ ions are better oxidizing agent than $Z n^{2 +}$ ions.
Similarly, $Z n$ is better reducing agent than $A u$ .

Suggest Corrections

Similar questions

Q. 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:

Q. A solution containing equimolar amounts of

N i C l_{2}

and

S n B r_{2}

is electrolysed using a 9V battery and graphite electrodes. What are the first products formed?
Half reaction

N i^{2 +} (a q) + 2 e^{-} \to N i (s)

Standard reduction potential (V) -0.236
Half reaction

S n^{2 +} (a q) + 2 e^{-} \to S n (s)

Standard reduction potential (V) -0.141

Half reaction

B r_{2} (a q) + 2 e^{-} \to 2 B r^{-} (a q)

Standard reduction potential (V) 1.077
Half reaction

C l_{2} (a q) + 2 e^{-} \to 2 C l^{-} (a q)

Standard reduction potential (V) 1.360

Q. The reaction :

{Z n}^{2 +} (a q) + {2 e}^{-} \to Z n (s)

has a electrode potential of

0.76 V

This means:

Q. Given standard electrode potential

F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.440 V

F e^{3 +} + 3 e^{-} \to F e; E^{0} = - 0.036

V. Calculate the electrode potential of cell.

Q. For the reduction of

N O_{3}^{-}

ion in an aqueous solution,

E^{o}

+ 0.96 V

. Values of

E^{o}

for some metal ions are given below.

V^{2 +} (aq) + 2 e^{-} \to V E^{0} = - 1.19 V

F e^{3 +} (aq) + 3 e^{-} \to F e E^{0} = - 0.04 V

A u^{3 +} (aq) + 3 e^{-} \to A u E^{0} = + 1.40 V

H g^{2 +} (aq) + 2 e^{-} \to H g E^{0} = + 0.86 V

The pair(s) of metals that is (are) oxidised by

N O_{3}^{-}

in aqueous solution is (are)

Join BYJU'S Learning Program

Electrode potentials are:Zn0→Zn2++2e−; E0=+0.76VAu0→Au3++3e−; E0=−1.42VWhat is the potential of the reaction, if a gold foil is kept in a solution containing zinc ions?

The correct option is C −2.18 V (no reaction)Eocell=EoAu−EoZn=−1.42−0.76=−2.18VSince, the cell potential is negative, the reaction is non spontaneous. Hence, no reaction will occur. Au3+ ions are better oxidizing agent than Zn2+ ions.Similarly, Zn is better reducing agent than Au.

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?