Question

Standard electrode potentials (\(\mathrm{M^{n+}/M}\)) of four metals A, B, C and D are given below: (A) \(-1.2\text{ V}\) (B) \(+0.5\text{ V}\) (C) \(+1.2\text{ V}\) (D) \(-3.0\text{ V}\) The correct increasing order for the reducing power of these metals will be:

• A. (D), (A), (C), (B)
• B. (D), (A), (B), (C)
• C. (C), (B), (A), (D)
• D. (B), (C), (A), (D)

Hint:

For metals, more negative standard reduction potential means stronger reducing agent because the metal is more easily oxidized.

Answer 1

The Correct Option is C

Concept: Reducing power of a metal depends upon its tendency to lose electrons. A metal having a more negative standard reduction potential \((E^\circ)\) is more readily oxidized and therefore acts as a stronger reducing agent. Thus, \[ \text{More negative } E^\circ \Rightarrow \text{Greater reducing power} \]

Step 1: Arrange the given electrode potentials.
\[ C=+1.2\text{ V} \] \[ B=+0.5\text{ V} \] \[ A=-1.2\text{ V} \] \[ D=-3.0\text{ V} \]

Step 2: Relate electrode potential to reducing nature.
Metal C has the highest positive reduction potential, therefore it has the least tendency to lose electrons and hence the least reducing power. Metal D has the most negative reduction potential and therefore the greatest tendency to lose electrons. Hence reducing power increases in the order: \[ C < B < A < D \]

Step 3: Write the required increasing order.
\[ \boxed{(C),(B),(A),(D)} \]