Question

The standard reduction potential at 25°C for \( MnO_4^- / H^+ \) is \( +1.49 \, \text{V} \).

The given standard reduction potentials are:

• \( Co^{3+}/Co^{2+} = +1.81 \, \text{V} \)
• \( Cr^{3+}/Cr = -0.74 \, \text{V} \)
• \( Au^{3+}/Au = +1.50 \, \text{V} \)
• \( Ag^+/Ag = +0.80 \, \text{V} \)

We need to identify which species cannot be oxidized by \( MnO_4^- / H^+ \).

Concept: A species will be oxidized by \( MnO_4^- \) only if its reduction potential is less than \( +1.49 \, \text{V} \).

Thus, species with higher reduction potential than \( MnO_4^- \) cannot be oxidized.

Comparing values:

• \( Co^{3+}/Co^{2+} = +1.81 \, \text{V} \) → cannot be oxidized
• \( Au^{3+}/Au = +1.50 \, \text{V} \) → cannot be oxidized

Final Answer: \( Co^{3+}/Co^{2+} \) and \( Au^{3+}/Au \)

• A. a & d
• B. a & c
• C. b & c
• D. b & d

Hint:

The ability of a substance to be oxidized depends on the relative reduction potentials. The lower the reduction potential, the easier it is to oxidize the substance.

Answer 1

The Correct Option is B

Step 1: Understand the oxidation process.
Oxidation involves loss of electrons. A species can be oxidized by \( MnO_4^- / H^+ \) only if its reduction potential is lower than that of \( MnO_4^- / H^+ \).
Higher reduction potential ⇒ greater tendency to be reduced ⇒ less likely to be oxidized.

Step 2: Given reduction potentials.
- \( Co^{3+}/Co^{2+} = +1.81 \, \text{V} \)
- \( Cr^{3+}/Cr = -0.74 \, \text{V} \)
- \( Au^{3+}/Au = +1.50 \, \text{V} \)
- \( Ag^+/Ag = +0.80 \, \text{V} \)

Step 3: Compare with \( MnO_4^- / H^+ \).
Given: \( E^\circ (MnO_4^- / H^+) = +1.49 \, \text{V} \)

Species with higher \( E^\circ \) than this cannot be oxidized.

- \( Co^{3+}/Co^{2+} \) ( +1.81 V ) → cannot be oxidized
- \( Au^{3+}/Au \) ( +1.50 V ) → cannot be oxidized

Step 4: Conclusion.
Thus, the species which cannot be oxidized are:
\( Co^{3+}/Co^{2+} \) and \( Au^{3+}/Au \)