JEE Chemistry 2026 Question with Solution

Given: The standard reduction potentials are

$$Al^{3+} + 3e^- \longrightarrow Al(s), \quad E^\circ = -1.66\,V$$ $$Cr^{3+} + 3e^- \longrightarrow Cr(s), \quad E^\circ = -0.74\,V$$ $$Fe^{3+} + e^- \longrightarrow Fe^{2+}, \quad E^\circ = +0.77\,V$$ $$Co^{3+} + e^- \longrightarrow Co^{2+}, \quad E^\circ = +1.81\,V$$

Find: The decreasing order of tendency to act as reducing agent.

A species acts as a stronger reducing agent if it has a greater tendency to undergo oxidation. This corresponds to a more negative standard reduction potential.

Comparing the given values:

$$-1.66 < -0.74 < +0.77 < +1.81$$

Therefore, the reducing strength decreases as:

$$Al > Cr > Fe^{2+} > Co^{2+}$$

Hence, the correct option is C.

Given: Standard reduction potentials of the corresponding couples are provided.

Find: Which species among $Al$, $Cr$, $Fe^{2+}$, and $Co^{2+}$ is the stronger reducing agent in decreasing order.

A reducing agent donates electrons, so it itself gets oxidized. If the reduction potential of a couple is more negative, the reverse oxidation process is more favorable.

Thus:

1. $Al$ is the strongest reducing agent because $E^\circ(Al^{3+}/Al) = -1.66\,V$ is the most negative.
2. $Cr$ comes next because $E^\circ(Cr^{3+}/Cr) = -0.74\,V$.
3. $Fe^{2+}$ is weaker because the couple $Fe^{3+}/Fe^{2+}$ has positive reduction potential $+0.77\,V$.
4. $Co^{2+}$ is the weakest reducing agent because $E^\circ(Co^{3+}/Co^{2+}) = +1.81\,V$ is the most positive.

So the required order is:

$$Al > Cr > Fe^{2+} > Co^{2+}$$

Therefore, the correct option is C.