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Question
Balance the following equations by the oxidation number method.
\[\ce{Fe^{2+} + H^{+} + Cr2O^{2-}7 -> Cr^{3+} + Fe^{3+} + H2O}\]
Solution
We can balance the given equation by oxidation number method-
\[\ce{\overset{+2}{Fe}^{2+} + H^{+} + \overset{+6}{C}r2\overset{-2}{O^{2-}7} -> \overset{+3}{C}r^{3+} + Fe^{3+} + H2O}\]
(a) Balance the increase and decrease in O.N.
\[\ce{6\overset{+2}{Fe}^{2+} + H^{+} + \overset{+6}{C}r2\overset{-2}{O^{2-}7} -> \overset{+3}{C}r^{3+} + 6\overset{3+}{F}e^{3+} + H2O}\]
(b) Balancing H and O atoms by adding H+ and H2O molecules.
\[\ce{6\overset{+2}{Fe}^{2+} + 14H^{+} + \overset{+6}{C}r2O^{2-}7 -> 2Cr^{3+} + 6Fe^{3+} + 7H2O}\]
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