What would be the coefficient of H+ ion in the balanced net ionic equations for the following reactions in acidic solution: S2O2−3(aq)+Cr2O2−7(aq)→S4O2−6(aq)+Cr3+(aq)
A
10
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B
8
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C
14
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D
12
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Solution
The correct option is C 14 Steps for Balancing redox reactions:
Identify the oxidation and reduction half .
Find the oxidising and reducing agent.
Find the n-factor of oxidising and reducing agent.
Balance atom undergoing oxidation and reduction.
Cross multiply the oxidising or reducing agent with n-factor.
Balance atoms other than oxygen and hydrogen.
Balancing oxygen atoms
Balancing hydrogen atoms
Balance charge
For acidic medium:
As soon as we add x H2O units, we add 2x H+ ions on the opposite side.
nf=(|O.S.Product−O.S.Reactant|×number of atom +2S2O2−3(aq)++6Cr2O2−7(aq)→+2.5S4O2−6(aq)++3Cr3+(aq)
+2S2O2−3(s)→+2.5S4O2−6(aq) oxidation nf=(|+2.5−2|×2=1 +6Cr2O2−7(aq)→+3Cr3+(aq) reduction nf=(|3−6|×2=6 Cr2O2−7 is oxidising agent S2O2−3 is reducing agent.
Balance atom undergoing oxidation and reduction. 2S2O2−3(aq)+Cr2O2−7(aq)→S4O2−6(aq)+2Cr3+(aq)
Cross mutiply the oxidising or reducing agent with n-factor. 12S2O2−3(aq)+Cr2O2−7(aq)→6S4O2−6(aq)+2Cr3+(aq)
Balance hydrogen. 12S2O2−3(aq)+Cr2O2−7(aq)+14H+→6S4O2−6(aq)+2Cr3+(aq)+7H2O
Balance charge
charge in reactant side = -12
charge in product side = -6
so the balanced equation is 12S2O2−3(aq)+Cr2O2−7(aq)+14H+→6S4O2−6(aq)+2Cr3+(aq)+7H2O+6e−
the coefficient of H+ is 14.