oxidizing agent?

Question

Examine the following half-reactions and select the weakest oxidizing agent among the species listed.
AuBr4-(aq) + 3e- ↔ Au(s) + 4Br-(aq) E° = 0.854 V
Mn2+(aq) + 2e- ↔ Mn(s) E° = -1.185 V
K+(aq) + e- ↔ K(s) E° = -2.931 V
F2O(aq) + 2H+(aq) + 4e- ↔ 2F-(aq) + H2O(l) E° = 2.153 V

AuBr4-(aq)
Mn2+(aq)
K+(aq)
F2O(aq)
H+(aq)

Answer 1

K+, which is the most negative.

Answer 2

I disagree. The stronger the oxidizing agent, the more negative the voltage of its redox half-reaction. I would use a Standard Reductions Potentials table to look at all the half-reactions for the elements involved in this question. whichever one has the biggest voltage is the strongest reducing agent and therefore the weakest oxidizing agent. From that, I got that the weakest oxidizing agent was F2O because it had the largest voltage (2.87V) on the Standard Reduction Potential table.