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

To determine the strongest reducing agent from the given options, we must understand what a reducing agent is. A reducing agent is a substance that loses electrons and gets oxidized. Therefore, among the given reduction half-reactions, the species that is most likely to lose electrons (i.e., the one that is least willing to undergo reduction) when reversing the reaction will be the strongest reducing agent.

To identify the strongest reducing agent, we need to look at the standard reduction potentials (\(E^0\)). The more negative the reduction potential, the stronger the reducing nature of the element or compound. Given potentials are:

• \(E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}\)
• \(E^0_{Cl_2/Cl^-} = 1.36 \text{ V}\)
• \(E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}\)
• \(E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}\)

To determine the reducing strength, we need to compare the standard reduction potentials of the following possible reducing agents:

• \(Cr\): The reduction potential is \(E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}\)
• \(Cl^-\): The reduction potential is \(E^0_{Cl_2/Cl^-} = 1.36 \text{ V}\) (reverse as Cl^- is oxidized from Cl₂)
• \(Mn^{2+}\): The MnO\(_4^-\) reduction to Mn\(^{2+}\) is \(E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}\) (reverse as Mn²⁺ is oxidized from MnO_4^-)

Among the potential reducing agents, \(Cr\) has the most negative standard reduction potential.

Conclusion: The element \(Cr\) is the strongest reducing agent among the given options because it has the most negative reduction potential at \(-0.74 \text{ V}\), making it more likely to get oxidized and act as a better reducing agent.