Standard reduction potentials are 1.455 V for the PbO2(s)/Pb(s) couple, 1.82 V for Co3 (aq)/Co2 (aq), 3.06 V for F2(g)/HF(aq), 1.07 V for Br2(l)/Br-(aq), and 1.77 V for H2O2(aq)/H2O(l). Under standard-state conditions, arrange the oxidizing agents in order of decreasing strength.

F2(g)/HF(aq)>Co3 (aq)/Co2 (aq)> H2O2(aq)/H2O(l)> PbO2(s)/Pb(s)>Br2(l)/Br-(aq)

Explanation:

The tendency of any specie to function as oxidizing agent is a highly dependent on the reduction potential of the couple. The more positive the value of the reduction potential of the couple, the better it does as an oxidizing agent.

This implies that we could know a good oxidizing agent by looking at their respective reduction potentials. The couple having the greatest (most positive) reduction potential is selected as the best oxidizing agent. If there are a number of couples at having different reduction potentials, the order of oxidizing ability can be obtained by arranging the species in order of decreasing positive reduction potentials just as we have done in the answer above.

a

the answer is d.) potassium