The strongest reducing agent amongst the following is:

(Given: \(E^0_{Ni^{2+}/Ni} = -0.25 V\), \(E^0_{Cu^{2+}/Cu} = 0.34 V\), \(E^0_{Pb^{2+}/Pb} = -0.13 V\), \(E^0_{Cr^{2+}/Cr} = -0.74 V\))

\(Cr^{2+}\)

The correct answer is option 4. \(Cr^{3+}\).

The strength of a reducing agent is determined by its tendency to lose electrons and get oxidized. In electrochemistry, this tendency is quantified by the standard reduction potential (\(E^0\)).

Understanding Standard Reduction Potential (\(E^0\)):

Reduction potential*measures the tendency of a species to gain electrons (be reduced). The more positive the \(E^0\), the greater the tendency to gain electrons (be reduced). Conversely, the more negative the \(E^0\), the greater the tendency for the species to lose electrons (be oxidized), making it a stronger reducing agent.

Given Data:

\(E^0_{Ni^{2+}/Ni} = -0.25 \, V\)

\(E^0_{Cu^{2+}/Cu} = 0.34 \, V\)

\(E^0_{Pb^{2+}/Pb} = -0.13 \, V\)

\(E^0_{Cr^{2+}/Cr} = -0.74 \, V\)

Analysis:

Copper (Cu): \(E^0_{Cu^{2+}/Cu} = 0.34 \, V\)

Positive value means copper is more likely to gain electrons (reduced). It is a poor reducing agent as it doesn’t easily lose electrons.

Lead (Pb): \(E^0_{Pb^{2+}/Pb} = -0.13 \, V\)

Slightly negative, indicating some tendency to lose electrons, but not very strong.

Nickel (Ni): \(E^0_{Ni^{2+}/Ni} = -0.25 \, V\)

More negative than lead, meaning it has a stronger tendency to lose electrons than lead.

Chromium (Cr): \(E^0_{Cr^{2+}/Cr} = -0.74 \, V\)

The most negative of all, indicating that chromium has the strongest tendency to lose electrons and get oxidized.

Conclusion:

Since \(Cr^{2+}/Cr\) has the most negative standard reduction potential (\(-0.74 \, V\)), chromium is the strongest reducing agent among the given options. It is the most likely to undergo oxidation (lose electrons), making it the most powerful in reducing other species.

Thus, the correct answer is Option 4 (\(Cr^{2+}\)).