Given two reactions and their electrode potential values-Reaction 1 -A aq- An- aq-ne-E-2 VReaction 2-B aq- Bn- aq-ne-E-4 VWhich of the following options is correct-

For nbsp;Reaction 1 : A (aq.)⇌ A^n+ (aq.)+ne^ E^∘=+2 V For reaction 2: B (aq.)⇌ B^n+ (aq.)+ne^ E^∘=+4 V Since ,E^∘ gt;0 for both the reactions, the forw ...

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Standard XII

Chemistry

Electrochemical Series

Given two rea...

Question

Given two reactions and their electrode potential values:
Reaction 1 :
$A (a q .) ⇌ A^{n +} (a q .) + n e^{-}$
$E^{\circ} = + 2 V$

Reaction 2:
$B (a q .) ⇌ B^{n +} (a q .) + n e^{-}$
$E^{\circ} = + 4 V$

Which of the following options is correct?

The reactant A has higher tendency to gain electrons than reactant B

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The reactant B has higher tendency to gain electrons than reactant A

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Tendency to gain electrons does not depend upon

E^{\circ}

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None of the above

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Solution

The correct option is A The reactant A has higher tendency to gain electrons than reactant B
For Reaction 1 :
$A (a q .) ⇌ A^{n +} (a q .) + n e^{-}$
$E^{\circ} = + 2 V$

For reaction 2:
$B (a q .) ⇌ B^{n +} (a q .) + n e^{-}$
$E^{\circ} = + 4 V$

Since , $E^{\circ} > 0$ for both the reactions, the forward reactions for both the reactants are spontaneous.
Higher the electrode potential, higher is the tendency for the forward reaction to take place .
Consequently higher is the tendency to lose electrons.
Since, for reaction 1, $E^{\circ}$ value is less than $E^{\circ}$ value of reaction 2, reactant A will have a higher tendency to gain electrons as compared to reactant B

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Similar questions

Q. Consider two reactions given below:
Reaction 1 :

A (a q .) ⇌ A^{n +} (a q .) + n e^{-}

E^{\circ} = 1.88 V

Reaction 2:

B (a q .) ⇌ B^{n +} (a q .) + n e^{-}

E^{\circ} = - 2.88 V

Which of the following options is correct?

Q. Given three reactions:
Reaction 1 :

A (a q .) ⇌ A^{n +} (a q .) + n e^{-}

E^{\circ} = + 2 V

Reaction 2:

B (a q .) ⇌ B^{n +} (a q .) + n e^{-}

E^{\circ} = - 3 V

Reaction 3:

C^{n +} (a q .) + n e^{-} ⇌ C (a q .)

E^{\circ} = + 4 V

The correct decreasing order of the tendency of the species A, B and C to get oxidised is:

Q. For a given electrochemical cell:

Given that:

A (s) ⇌ A^{n +} (a q) + n e^{-}

E^{\circ} = - 1.5 V

B (s) ⇌ B^{n +} (a q) + n e^{-}

E^{\circ} = - 3.1 V

Choose the correct statement.

Q. Use the following

E^{o}

for the electrode potentials, calculate

Δ G^{o}

k J

for the indicated reaction.

5 C e^{4 +} (a q) + M n^{2 +} (a q) + 4 H_{2} O (l) \to M n O_{4}^{-} (a q) + 8 H^{+} (a q) M n O_{4}^{-} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l); E^{o} = + 1.51 V C e^{4 +} + e^{-} \to C e^{3 +} (a q); E^{o} = + 1.61 V

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox Titration. Some half – cell reactions and their standard potential are given below:
$M n O_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (I); E^{\circ} = 1.51 V$
$C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O (I); E^{\circ} = 1.38 V$
$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{\circ} = 0.77 V$
$C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q); E^{\circ} = 1.40 V$

Identify the only incorrect statement regarding the quantitative estimation of aqueous $F e (N O_{3})_{2}$ .

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Given two reactions and their electrode potential values: Reaction 1 : A (aq.)⇌An+ (aq.)+ne− E∘=+2 V Reaction 2: B (aq.)⇌Bn+ (aq.)+ne− E∘=+4 V Which of the following options is correct?

Given two reactions and their electrode potential values:
Reaction 1 :
$A (a q .) ⇌ A^{n +} (a q .) + n e^{-}$
$E^{\circ} = + 2 V$

Reaction 2:
$B (a q .) ⇌ B^{n +} (a q .) + n e^{-}$
$E^{\circ} = + 4 V$

Which of the following options is correct?