Question

If standard reduction potential of Zn, Ni and Fe are \(-0.76\,\text{V}\), \(-0.23\,\text{V}\) and \(-0.44\,\text{V}\) respectively. For the reaction (Stated below) to be spontaneous find out electrodes X and Y considering above electrode potentials
\(\text{X}_{(s)}+\text{Y}_{(aq)}^{+2}⟶\text{X}_{(aq)}^{+2}+\text{Y}_{(s)}\)

• A. X = Ni , Y = Fe
• B. X = Ni , Y = Zn
• C. X = Fe , Y = Zn
• D. X = Zn , Y = Ni

Hint:

A reaction is spontaneous when X (oxidised) has a lower reduction potential than Y (reduced).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In the reaction \(\text{X} + \text{Y}^{2+} \to \text{X}^{2+} + \text{Y}\), X is oxidised (anode) and \(\text{Y}^{2+}\) is reduced (cathode).

Step 2: Key Formula or Approach:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} > 0 \]
So the reduced species (Y) must have a higher reduction potential than the oxidised species (X).

Step 3: Detailed Explanation:
Reduction potentials: Zn -0.76 V, Fe -0.44 V, Ni -0.23 V.
Check (A) X = Ni, Y = Fe: \(E^{\circ} = -0.44 - (-0.23) = -0.21\) V, not spontaneous.
Check (B) X = Ni, Y = Zn: \(E^{\circ} = -0.76 + 0.23 = -0.53\) V, not spontaneous.
Check (C) X = Fe, Y = Zn: \(E^{\circ} = -0.76 + 0.44 = -0.32\) V, not spontaneous.
Check (D) X = Zn, Y = Ni: \(E^{\circ} = -0.23 - (-0.76) = +0.53\) V, positive, so spontaneous.

Final Answer:
Zinc is oxidised and nickel ion is reduced, so X = Zn and Y = Ni, option (D). \[ \boxed{X=\text{Zn},\ Y=\text{Ni} \text{ (D)}} \]