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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?

Respuesta :

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.