Question

The standard electrode potentials for \(\text{Zn}^{2+}/\text{Zn}\), \(\text{Ni}^{2+}/\text{Ni}\), and \(\text{Fe}^{2+}/\text{Fe}\) are \(-0.76\text{ V}\), \(-0.23\text{ V}\), and \(-0.44\text{ V}\) respectively. Which of the following reactions is spontaneous?

• A. \(\text{Ni}^{2+} + \text{Fe} \rightarrow \text{Ni} + \text{Fe}^{2+}\)
• B. \(\text{Ni} + \text{Zn}^{2+} \rightarrow \text{Ni}^{2+} + \text{Zn}\)
• C. \(\text{Fe} + \text{Zn}^{2+} \rightarrow \text{Fe}^{2+} + \text{Zn}\)
• D. \(\text{Zn}^{2+} + \text{Fe}^{2+} \rightarrow \text{Zn} + \text{Fe}\)

Hint:

A metal with a more negative standard reduction potential acts as a stronger reducing agent and will displace a metal with a less negative reduction potential from its aqueous salt solution. (Zn displaces Fe and Ni; Fe displaces Ni).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to determine which of the provided redox reactions is thermodynamically spontaneous based on the standard reduction potentials.

Step 2: Key Formula or Approach:
A reaction is spontaneous if the standard cell potential (\(E^\circ_{\text{cell}}\)) is positive (\(E^\circ_{\text{cell}} > 0\)).
The formula for standard cell potential is:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] where the cathode undergoes reduction and the anode undergoes oxidation.

Step 3: Detailed Explanation:
The given standard reduction potentials are:
\(E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}\)
\(E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44\text{ V}\)
\(E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.23\text{ V}\)
Let's evaluate \(E^\circ_{\text{cell}}\) for each provided option:

(A) \(\text{Ni}^{2+} + \text{Fe} \rightarrow \text{Ni} + \text{Fe}^{2+}\)
Reduction (Cathode): \(\text{Ni}^{2+} \rightarrow \text{Ni}\) \(\implies E^\circ = -0.23\text{ V}\)
Oxidation (Anode): \(\text{Fe} \rightarrow \text{Fe}^{2+}\) \(\implies E^\circ = -0.44\text{ V}\)
\[ E^\circ_{\text{cell}} = (-0.23\text{ V}) - (-0.44\text{ V}) = +0.21\text{ V} \] Since \(E^\circ_{\text{cell}} > 0\), this reaction is

spontaneous.

(B) \(\text{Ni} + \text{Zn}^{2+} \rightarrow \text{Ni}^{2+} + \text{Zn}\)
Reduction (Cathode): \(\text{Zn}^{2+} \rightarrow \text{Zn}\) \(\implies E^\circ = -0.76\text{ V}\)
Oxidation (Anode): \(\text{Ni} \rightarrow \text{Ni}^{2+}\) \(\implies E^\circ = -0.23\text{ V}\)
\[ E^\circ_{\text{cell}} = (-0.76\text{ V}) - (-0.23\text{ V}) = -0.53\text{ V} \] Since \(E^\circ_{\text{cell}} < 0\), this reaction is non-spontaneous.

(C) \(\text{Fe} + \text{Zn}^{2+} \rightarrow \text{Fe}^{2+} + \text{Zn}\)
Reduction (Cathode): \(\text{Zn}^{2+} \rightarrow \text{Zn}\) \(\implies E^\circ = -0.76\text{ V}\)
Oxidation (Anode): \(\text{Fe} \rightarrow \text{Fe}^{2+}\) \(\implies E^\circ = -0.44\text{ V}\)
\[ E^\circ_{\text{cell}} = (-0.76\text{ V}) - (-0.44\text{ V}) = -0.32\text{ V} \] Since \(E^\circ_{\text{cell}} < 0\), this reaction is non-spontaneous.

(D) \(\text{Zn}^{2+} + \text{Fe}^{2+} \rightarrow \text{Zn} + \text{Fe}\)
This represents two simultaneous reductions without an oxidation half-reaction, which is chemically impossible in a simple redox cell.

Step 4: Final Answer:
The correct choice is (A).