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Question

What will be the balanced equation in acidic medium for the given reaction ?
Cr2O27(aq)+SO2(g)Cr3+(aq)+SO24(aq)

A
Cr2O27(aq)+3SO2(g)+2H+(aq)2Cr3+(aq)+3SO24(aq)+H2O(l)
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B
2Cr2O27(aq)+3SO2(g)+4H+(aq)4Cr3+(aq)+3SO24(aq)+2H2O(l)
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C
Cr2O27(aq)+3SO2(g)+14H+(aq)2Cr3+(aq)+3SO24(aq)+7H2O(l)
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D
Cr2O27(aq)+6SO2(g)+7H+(aq)2Cr3+(aq)+6SO24(aq)+7H2O(l)
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Solution

The correct option is A Cr2O27(aq)+3SO2(g)+2H+(aq)2Cr3+(aq)+3SO24(aq)+H2O(l)
In a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.
Balancing a chemical reaction as:
Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:
+6Cr2O72(aq)++4SO2+3Cr++6SO24
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: +6Cr2O72(aq)2+3Cr
Oxidation: +4SO2+6SO24
Step 3: oxidation-number change is:
Reduction: +6Cr2O722+3Cr: Gain of 6 electrons total
Oxidation: +4SO2+6SO24 Loss of total 2 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: +6Cr2O722+3Cr×1: Gain of 6 electron total
Oxidation: +4SO2+6SO24×3 : Loss of 6 electron
Reduction: +6Cr2O722+3Cr
Oxidation: 3+4SO23+6SO24
Step 5: Balance O atoms in reduction reaction by adding H2O and then balance H by H+ as:
Reduction: +6Cr2O72+14H+2+3Cr+7H2O
Oxidation: 3+4SO2+6H2O3+6SO24+12H+
Thus overall Balanced reaction is:
+6Cr2O72+14H++3+4SO2+6H2O2+3Cr+7H2O+3+6SO24+12H+
or +6Cr2O72+2H++3+4SO22+3Cr+H2O+3+6SO24

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