Question

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?

Answer 1

Answer:

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.