Question

If standard reduction potential \((E^0)\) of \((\text{Al}_{(aq)}^{+3}|\text{Al}_{(s)})\), \((\text{Fe}_{(aq)}^{+2}|\text{Fe}_{(s)})\), \((\text{Cu}_{(aq)}^{+2}|\text{Cu}_{(s)})\) and \((\text{Ag}_{(aq)}^{+1}|\text{Ag}_{(s)})\) are \(-1\cdot 66\) V, \(-0\cdot 44\) V, \(+0\cdot 34\) V and \(+0\cdot 79\) V respectively. Which of the following reaction is non spontaneous ?

• A. \(2\text{Ag}_{(s)}+\text{Fe}_{(aq)}^{+2}\rightarrow 2\text{Ag}_{(aq)}^{+1}+\text{Fe}_{(s)}\)
• B. \(2\text{Al}_{(s)}+3\text{Cu}_{(aq)}^{+2}\rightarrow 2\text{Al}_{(aq)}^{+3}+3\text{Cu}_{(s)}\)
• C. \(\text{Fe}_{(s)}+\text{Cu}_{(aq)}^{+2}\rightarrow \text{Fe}_{(aq)}^{+2}+\text{Cu}_{(s)}\)
• D. \(2\text{Al}_{(s)}+3\text{Fe}_{(aq)}^{+2}\rightarrow 2\text{Al}_{(aq)}^{+3}+3\text{Fe}_{(s)}\)

Hint:

A reaction is spontaneous when the cell EMF, E(cathode) minus E(anode), is positive.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A redox reaction is spontaneous if \(E^0_{cell} = E^0_{reduction\ (cathode)} - E^0_{reduction\ (anode)} > 0\).

Step 2: Check (A):
\(2\text{Ag} + \text{Fe}^{2+} \rightarrow 2\text{Ag}^+ + \text{Fe}\). Silver is oxidised (anode, 0.79 V) and \(\text{Fe}^{2+}\) is reduced (cathode, -0.44 V). \(E^0_{cell} = -0.44 - 0.79 = -1.23\) V. This is negative, so (A) is NON spontaneous.

Step 3: Check (B), (C), (D):
(B) Al anode, Cu cathode: \(0.34 - (-1.66) = +2.00\) V, spontaneous.
(C) Fe anode, Cu cathode: \(0.34 - (-0.44) = +0.78\) V, spontaneous.
(D) Al anode, Fe cathode: \(-0.44 - (-1.66) = +1.22\) V, spontaneous.

Final Answer:
Only reaction (A) has a negative cell potential, so it is non spontaneous. \[ \boxed{\text{(A)}} \]