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Question

Write balanced net ionic equation for the following reactions in acidic solution.
S2O62(aq)+Cr2O72(aq)S4O62(aq)+Cr3+(aq)

A
6S2O62(aq)+Cr2O72(aq)+14H+3S4O62(aq)+2Cr3+(aq)+7H2O
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B
6S2O62(aq)+2Cr2O72(aq)+14H+3S4O62(aq)+2Cr3+(aq)+7H2O
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C
6S2O62(aq)+Cr2O72(aq)+14H+4S4O62(aq)+2Cr3+(aq)+7H2O
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D
None of the above
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Solution

The correct option is A 6S2O62(aq)+Cr2O72(aq)+14H+3S4O62(aq)+2Cr3+(aq)+7H2O
The unbalanced net ionic equation is S2O62(aq)+Cr2O72(aq)S4O62(aq)+Cr3+(aq)

Balance S atoms and Cr atoms
2S2O62(aq)+Cr2O72(aq)S4O62(aq)+2Cr3+(aq)

The oxidation number of Cr decreases from +6 to +3. Total decrease in the oxidation number for 2Cr atoms is 6.

To balance the decrease in the oxidation number with increase in the oxidation number, multiply sulphur species with 3.
6S2O62(aq)+Cr2O72(aq)3S4O62(aq)+2Cr3+(aq)

To balance O atoms, add 7 water molecules on RHS
6S2O62(aq)+Cr2O72(aq)3S4O62(aq)+2Cr3+(aq)+7H2O

To balance hydrogen atoms, add 14 protons on the LHS.
6S2O62(aq)+Cr2O72(aq)+14H+3S4O62(aq)+2Cr3+(aq)+7H2O

This is balanced chemical equation

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