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Question
Balance the following reaction by oxidation number method.
\[\ce{Cr2O^2-_{7(aq)} + SO^2-_{3(aq)}->Cr^3+_{ (aq)} + SO^2-_{4(aq)}(acidic)}\]
Solution
\[\ce{Cr2O^2-_{7(aq)} + SO^2-_{3(aq)}->Cr^3+_{ (aq)} + SO^2-_{4(aq)}(acidic)}\]
Step 1: Write the skeletal equation and balance the elements other than O and H.
\[\ce{Cr2O^2-_{7(aq)} + SO^2-_{3(aq)}->2Cr^3+_{ (aq)} + SO^2-_{4(aq)}}\]
Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.
Increase in oxidation number:
(Increase per atom = 2)
Decrease in oxidation number:
(Decrease per atom = 3)
To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2Cr atoms.)
\[\ce{Cr2O^2-_{7(aq)} + 3SO^2-_{3(aq)}->2Cr^3+_{ (aq)} + 3SO^2-_{4(aq)}}\]
Step 3: Balance 'O' atoms by adding 4H2O to the right-hand side.
\[\ce{Cr2O^2-_{7(aq)} + 3SO^2-_{3(aq)}->2Cr^3+_{ (aq)} + 3SO^2-_{4(aq)} + 4H2O_{(l)}}\]
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H+ on the left-hand side.
\[\ce{Cr2O^2-_{7(aq)} + 3SO^2-_{3(aq)} + 8H^+_{ (aq)}->2Cr^3+_{ (aq)} + 3SO^2-_{4(aq)} + 4H2O_{(l)}}\]
Step 5: Check two sides for the balance of atoms and charges.
Hence, balanced equation: \[\ce{Cr2O^2-_{7(aq)} + 3SO^2-_{3(aq)} + 8H^+_{ (aq)}->2Cr^3+_{ (aq)} + 3SO^2-_{4(aq)} + 4H2O_{(l)}}\]
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