Standard reduction electrode potential of four metals A, B, C and D are as follows. The correct order of increasing strength of reducing power is:

A. Metal A = +0.5 V

B. Metal B = -3.0 V

C. Metal C = -1.2 V

D. Metal D = -2.3 V

Choose the correct answer from the options given below:

A > C > D > B

The correct answer is option 3. A > C > D > B.

The standard reduction electrode potential (E°) of a metal is a measure of its tendency to gain electrons and be reduced. The more positive the E° value, the greater the tendency of the metal ion to gain electrons and undergo reduction. Conversely, the more negative the E° value, the lesser the tendency of the metal ion to be reduced, and instead, the metal tends to lose electrons and act as a reducing agent.

A reducing agent is a substance that donates electrons to another species and is itself oxidized in the process. The strength of a reducing agent is directly related to its ability to lose electrons. Therefore:

A metal with a more negative E° is a stronger reducing agent because it has a greater tendency to lose electrons (be oxidized).

A metal with a more positive E° is a weaker reducing agent because it has a greater tendency to gain electrons (be reduced).

The reduction potentials of the metals provided are:

Metal A: \( +0.5 \, \text{V} \)

Metal B: \( -3.0 \, \text{V} \)

Metal C: \( -1.2 \, \text{V} \)

Metal D: \( -2.3 \, \text{V} \)

Interpreting Each Metal's Reducing Power

Metal A \( (+0.5 \, \text{V}) \):

This is the most positive E° value among the metals listed. A positive E° value means Metal A has a strong tendency to gain electrons (be reduced) and a very weak tendency to lose electrons (be oxidized). Therefore, Metal A is the weakest reducing agent.

Metal C \( (-1.2 \, \text{V}) \):

This E° value is negative, indicating that Metal C has a lesser tendency to gain electrons and a greater tendency to lose electrons compared to Metal A. Metal C is a stronger reducing agent than Metal A, but weaker than metals with more negative E° values.

Metal D \( (-2.3 \, \text{V}) \):

With a more negative E° than Metal C, Metal D has an even stronger tendency to lose electrons and be oxidized. Metal D is a stronger reducing agent than both Metal A and Metal C.

Metal B \( (-3.0 \, \text{V}) \):

This is the most negative E° value, indicating that Metal B has the highest tendency to lose electrons (be oxidized). Metal B is the strongest reducing agent among the four metals.

To arrange the metals in order of increasing reducing power, we list them from the weakest reducing agent (the most positive E°) to the strongest reducing agent (the most negative E°):

Metal A \( (+0.5 \, \text{V}) \) is the weakest reducing agent.

Metal C \( (-1.2 \, \text{V}) \) is stronger than Metal A.

Metal D \( (-2.3 \, \text{V}) \) is stronger than Metal C.

Metal B \( (-3.0 \, \text{V}) \) is the strongest reducing agent.

Thus, the order of increasing reducing power is: \(\text{A > C > D > B} \)

Conclusion

The correct answer is option 3: A > C > D > B. This sequence represents the correct order of increasing strength of reducing power, from the weakest (A) to the strongest (B).