Arrange the following electrodes in increasing order of strength as an oxidizing agent.

(A) $Ni^{2+} + 2e^- → Ni(s)\,\,\,\,\, E° = -0.25 V$
(B) $Cu^{2+} + 2e^- → Cu(s)\,\,\,\,\, E° = 0.34 V$
(C) $Br_2 + 2e^- → 2Br\,\,\,\,\, E° = 1.09 V$
(D) $O_2(g) + 2H^+ + 2e^- → H_2O_2\,\,\,\,\, E° = 0.68 V$

Choose the correct answer from the options given below:

(A), (B), (D), (C)

The correct answer is Option (4) → (A), (B), (D), (C)

The strength of an oxidizing agent is directly related to its Standard Reduction Potential ($E^\circ$). An oxidizing agent is a substance that gains electrons and is itself reduced. Therefore, the higher (more positive) the reduction potential, the greater the tendency of the species to accept electrons and the stronger it acts as an oxidizing agent.

Step-by-Step Arrangement

To arrange the electrodes in increasing order of strength as an oxidizing agent, we simply list their $E^\circ$ values from the most negative to the most positive:

(A) $Ni^{2+} + 2e^- \rightarrow Ni(s)$: $E^\circ = -0.25\text{ V}$ (Lowest value, weakest oxidizing agent)

(B) $Cu^{2+} + 2e^- \rightarrow Cu(s)$: $E^\circ = +0.34\text{ V}$

(D) $O_2(g) + 2H^+ + 2e^- \rightarrow H_2O_2$: $E^\circ = +0.68\text{ V}$

(C) $Br_2 + 2e^- \rightarrow 2Br^-$: $E^\circ = +1.09\text{ V}$ (Highest value, strongest oxidizing agent)