Which is the strongest oxidizing agent? lithium; li+ + e– → li e0 = –3.05 v hydrogen; 2h+ + 2e– → h2e0 = 0.00 v fluorine; f2 + 2e– → 2f–e0 = +2.87 v?

Fluorine; f2 + 2e– → 2f–e0 = +2.87 v

Explanation:

Oxidizing agents remove electrons from other atoms to complete a stable outer octet. In this case, Flourine is an oxidizing agent.Hydrogen and lithium are reducing agents because they can lose electrons more easily than accept them. We can also use the electrode potential or e.m.f to determine whether an element is a strong oxidizing agent or not. An element with the largest positive e.m.f is the strongest oxidizing agent as it indicates that its more electronegative.

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They are all decomposition reactions.