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Question
Balance the following equation in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
\[\ce{Cl_2O_{7(g)} + H_2O_{2(aq)} -> ClO-_{2(aq)} + O_{2(g)} + H+_{(aq)}}\]
Solution
The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in `"ClO"_2^-` and the oxidation number of O increases from – 1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the oxidizing agent and H2O2 is the reducing agent.
Ion–electron method:
The oxidation half equation is:
\[\ce{H2 ^{-1}O_{2(g)} -> ^{0}O_{2(g)}}\]
The oxidation number is balanced by adding 2 electrons as:
\[\ce{H_2O_{2(aq)} -> O_{2(g)} + 2e-}\]
The charge is balanced by adding 2OH– ions as:
\[\ce{H_2O_{2(aq)} -> + 2OH-_{(aq)} -> O_{2(g)} + 2e-}\]
The oxygen atoms are balanced by adding 2H2O as:
\[\ce{H_2O_{2(aq)} + 2OH-_{(aq)} -> O_{2(g)} + 2H_2O_{(l)} + 2e-}\] ....(i)
The reduction half equation is:
\[\ce{^{+7}Cl2O_{7(g)} -> ^{+3}ClO-_{2(aq)}}\]
The Cl atoms are balanced as:
\[\ce{Cl2O_{7(g)} -> 2ClO-_{2(aq)}}\]
The oxidation number is balanced by adding 8 electrons as:
\[\ce{Cl2O_{7(g)} + 8e- -> 2ClO-_{2(aq)}}\]
The charge is balanced by adding 6OH– as:
\[\ce{Cl_2O_{7(g)} + 8e- -> 2ClO-_{2(aq)} + 6OH-_{(aq)}}\]
The oxygen atoms are balanced by adding 3H2O as:
\[\ce{Cl_2O_{7(g)} + 3H_2O_{(l)} 8e- -> 2ClO-_{2(aq)} + 6OH-_{(aq)}}\] (ii)
The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:
\[\ce{Cl_2O_{7(g)} + 4H_2O_{2(aq)} + 2OH-_{aq)} -> 2ClO-_{2(aq)} + 4O_{2(g)} + 5H_2O_{(l)}}\]
Oxidation number method:
Total decrease in oxidation number of Cl2O7 = 4 × 2 = 8
Total increase in oxidation number of H2O2 = 2 × 1 = 2
By multiplying H2O2 and O2 with 4 to balance the increase and decrease in the oxidation number, we get:
\[\ce{Cl_2O_{7(g)} + 4H_2O_{2(aq)} -> ClO-_{2(aq)} + 4O_{2(g)}}\]
The Cl atoms are balanced as:
\[\ce{Cl_2O_{7(g)} + 4H_2O_{2(aq)} -> 2ClO-_{2(aq)}+ 4O_{2(g)}}\]
The O atoms are balanced by adding 3H2O as:
\[\ce{Cl_2O_{7(g)} + 4H_2O_{2(aq)} -> 2ClO-_{2(aq)} + 4O_{2(g)} + 3H_2O_{(l)}}\]
The H atoms are balanced by adding 2OH– and 2H2O as:
\[\ce{Cl_2O_{7(g)} + 4H_2O_{2(aq)} + 2OH-_{(aq)} -> 2ClO-_{2(aq)} + 4O_{2(g)} + 5H_2O_{(l)}}\]
This is the required balanced equation.
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