Question 4.9: IDENTIFYING OXIDIZING AND REDUCING AGENTS Assign oxidation n...
IDENTIFYING OXIDIZING AND REDUCING AGENTS
Assign oxidation numbers to all atoms, tell in each case which substance is undergoing oxidation and which reduction, and identify the oxidizing and reducing agents.
(a) Ca(s) + 2 H^{+}(aq) → Ca^{2+}(aq) + H_{2}(g)
(b) 2 Fe^{2+}(aq) + Cl_{2}(aq) → 2 Fe^{3+}(aq) + 2 Cl^{-}(aq)
(a) The elements Ca and H_{2} have oxidation numbers of 0; Ca^{2+} is 2 and H^{+} is +1:
\underset{\overset{ \uparrow }{0} }{Ca}(s) + 2 \underset{\overset{ \uparrow }{+1} }{H^{+}}(aq) → \underset{\overset{ \uparrow }{+2} }{Ca^{2+}} (aq) + \underset{\overset{ \uparrow }{0} }{H_{2}} (g)
Ca is oxidized because its oxidation number increases from 0 to 2, and H^{+} is reduced because its oxidation number decreases from +1 to 0. The reducing agent is the substance that gives away electrons, thereby going to a higher oxidation number, and the oxidizing agent is the substance that accepts electrons, thereby going to a lower oxidation number. In the present case, calcium is the reducing agent and H^{+} is the oxidizing agent.
(b) Atoms of the neutral element Cl_{2} have an oxidation number of 0; the monatomic ions have oxidation numbers equal to their charge:
2 \underset{\overset{ \uparrow }{+2} }{Fe^{2+}}(aq) + \underset{\overset{ \uparrow }{0} }{Cl_{2}}(aq) → 2 \underset{\overset{ \uparrow }{+3} }{Fe^{3+}} (aq) + 2 \underset{\overset{ \uparrow }{-1} }{Cl^{-}} (aq)Fe^{2+} is oxidized because its oxidation number increases from +2 to +3, and Cl_{2}is reduced because its oxidation number decreases from 0 to -1. Fe^{2+} is the reducing agent, and Cl_{2} is the oxidizing agent.