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Question
Balance the following equation by the ion electron method.
\[\ce{Zn + NO^-_3 ->Zn^2+ + NO}\] (in acid medium)
Solution
Half reactions are
\[\ce{Z^0n -> Zn^2+}\] ...........(1)
\[\ce{N^{+5}O^-_3 -> N^2+O}\] ......(2)
\[\ce{Zn -> Zn^2+ + 2e^-}\] .........(3)
\[\ce{NO^-_3 + 3e^- + 4H^+ -> NO + 2H2O}\] ........(4)
\[\ce{3Zn -> 3Zn^2+ + 6e^-}\] .......(5) [Eq. (3) × 3]
\[\ce{2NO^-_3 + 6e^- + 8H^+ -> 2NO + 4H2O}\] ...........(6) [Eq. (4) × 2]
\[\ce{3Zn + 2NO^-_3 + 8H^+ -> 3Zn^2+ + 2NO + 4H2O}\]
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