For the equation in acidic medium: S2O2−3(aq)+Cr2O2−7(aq)→S4O2−6(aq)+Cr3+(aq) The coefficient of H+ in the balanced reaction by oxidation number method will be :
A
5
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B
7
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C
10
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D
14
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Solution
The correct option is D 14 +2S2O2−3(aq)++6Cr2O2−7(aq)→+2.5S4O2−6(aq)++3Cr3+(aq) Cr2O2−7 is oxidising agent. S2O2−3 is reducing agent.
Balancing elements except O and H 6S2O2−3(aq)+Cr2O2−7(aq)→3S4O2−6(aq)+2Cr3+(aq)
Balancing oxygen atoms by adding H2O 6S2O2−3(aq)+Cr2O2−7(aq)→3S4O2−6(aq)+2Cr3+(aq)+7H2O
Balancing hydrogen by adding H+ 6S2O2−3(aq)+Cr2O2−7(aq)+14H+→3S4O2−6(aq)+2Cr3+(aq)+7H2O
Balancing charge: charge in reactant side = -12-2+14=0 charge in product side = -6+6=0 so the balanced equation is 6S2O2−3(aq)+Cr2O2−7(aq)+14H+→3S4O2−6(aq)+2Cr3+(aq)+7H2O
The coefficient of H+ in balanced equation is is 14.
Theory : Balancing of redox reactions : For a redox reaction to be balanced : 1. Reaction should have Atom balance on both reactant and product side 2. Reaction should have Charge balance on both reactant and product side.
Oxidation Number method : (Acidic medium) Step 1: Identifying the oxidizing/reducing agent Step 2: find the nf by the formula nf=(|O.S.Product−O.S.Reactant|×number of atom Step 3: Equalising the decrease/increase in oxidation number Cross multiply the Oxidising and the Reducing Agents by each other respectively on the reactant side. Step 4: Balance all the atoms except O and H, without changing the stoichiometric coefficients on the reactant side Step 5: Balancing the O atoms Balance the O atoms by adding H2O Step 6: For an acidic medium : As soon as we add H2O, we add twice the H+ ions on the opposite side. xH2O⟶2xH+