Given standard reduction potentials

$E^\circ(\text{A}^{n+}/\text{A}) = +0.5\text{ V}$ $E^\circ(\text{B}^{n+}/\text{B}) = -3.0\text{ V}$ $E^\circ(\text{C}^{n+}/\text{C}) = -1.2\text{ V}$

Concept used: reducing power of a metal

A metal acts as a reducing agent by getting oxidized: $\text{M} \rightarrow \text{M}^{n+} + ne^-$

The stronger the tendency of the metal to undergo oxidation, the greater is its reducing power.

Since the given values are reduction potentials, a more negative standard reduction potential means:

• reduction is less favorable,
• oxidation of the metal is more favorable,
• hence the metal is a stronger reducing agent.

Compare the values

$E^\circ_B = -3.0\text{ V}$ $E^\circ_C = -1.2\text{ V}$ $E^\circ_A = +0.5\text{ V}$

Most negative potential corresponds to strongest reducing power.

Therefore, $\text{Reducing power: } B > C > A$

Match with options

• A: $A > B > C$ ❌
• B: $C > B > A$ ❌
• C: $A > C > B$ ❌
• D: $B > C > A$ ✅

Final answer

$\boxed{B > C > A}$ So, the correct option is D.