Question

Based on the data given below:
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$   $E^0_{Cl_2/Cl^-} = 1.36 \text{ V}$
$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$    $E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$
the strongest reducing agent is:

• A. \(Mn^{2+}\)
• B. \(Cr\)
• C. \(MnO_4^-\)
• D. \(Cl^-\)

Hint:

Reduction Potential \(\propto\) Oxidising Power.
Oxidation Potential \(\propto\) Reducing Power.
Since \(\text{Oxidation Potential} = -\text{Reduction Potential}\), the most negative \(E^0_{\text{red}}\) gives the highest reducing power.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A reducing agent is a species that loses electrons (undergoes oxidation).
The tendency of a species to be reduced is measured by its standard reduction potential (\(E^0\)).
The species with the lowest (most negative) \(E^0\) value has the greatest tendency to undergo oxidation and therefore acts as the strongest reducing agent.

Step 2: Key Formula or Approach:
1. List all given standard reduction potentials.
2. Identify the lowest \(E^0\) value.
3. The reduced form (the species on the right side of the reduction half-reaction) corresponding to the lowest \(E^0\) is the strongest reducing agent.

Step 3: Detailed Explanation:
Given potentials:
1. \(E^0(MnO_4^-/Mn^{2+}) = 1.51 \text{ V}\)
2. \(E^0(Cl_2/Cl^-) = 1.36 \text{ V}\)
3. \(E^0(Cr_2O_7^{2-}/Cr^{3+}) = 1.33 \text{ V}\)
4. \(E^0(Cr^{3+}/Cr) = -0.74 \text{ V}\)
The lowest value is \(-0.74 \text{ V}\) for the couple \(Cr^{3+}/Cr\).
This means \(Cr\) metal has the highest tendency to oxidize to \(Cr^{3+}\) among the given species.
Hence, \(Cr\) is the strongest reducing agent.

Step 4: Final Answer:
Based on the minimum reduction potential, \(Cr\) is the strongest reducing agent.