The correct option is
A Cr2O2−7(aq)+3SO2(g)+2H+(aq)→2Cr3+(aq)+3SO2−4(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++6SO2−4
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: +6Cr2O72−(aq)→2+3Cr
Oxidation: +4SO2→+6SO2−4
Step 3: oxidation-number change is:
Reduction: +6Cr2O72−→2+3Cr: Gain of 6 electrons total
Oxidation: +4SO2→+6SO2−4 Loss of total 2 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: +6Cr2O72−→2+3Cr×1: Gain of 6 electron total
Oxidation: +4SO2→+6SO2−4×3 : Loss of 6 electron
∴ Reduction: +6Cr2O72−→2+3Cr
Oxidation: 3+4SO2→3+6SO2−4
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+6H2O→3+6SO2−4+12H+
Thus overall Balanced reaction is:
+6Cr2O72−+14H++3+4SO2+6H2O→2+3Cr+7H2O+3+6SO2−4+12H+
or +6Cr2O72−+2H++3+4SO2→2+3Cr+H2O+3+6SO2−4