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Question

balance the redox reactions oxidations number method /Half reactions method
1. H2O 2(aq) +Fe3+ (aq)+----Fe3+ (aq)+ H2O(l) (in acidic solution)

2.cr2O72- +SO 2 (g)---cr 3+ (aq)+ so42-(aq) (in acidic solution)

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Solution

Dear Student,

1. H2O2 + Fe2+ Fe3+ + H2ODivide the equation into two half cellsOxidation : Fe2+ Fe3+ Reduction : H2O2 H2OBalance all atoms other than H and OOxidation : Fe2+ Fe3+ Reduction : H2O2 H2OBalance O by adding H2O to O deficient sideOxidation : Fe2+ Fe3+ Reduction : H2O2 H2O + H2OBalance H by adding H+ to H deficient sideOxidation : Fe2+ Fe3+ Reduction : H2O2 +2H+ 2H2OBalance charge by adding electronsOxidation : Fe2+ Fe3+ +e- Reduction : H2O2 +2H++e- 2H2OMultiply the oxidation half by 2 and add both equations2Fe2+ H2O2 +2H+ 2Fe3+ +2H2OThis is the required balanced equation. 2. Cr2O72-+ SO2 Cr3+ + SO42-Divide the equation into two half cellsOxidation : SO2 SO42-Reduction : Cr2O72- Cr3+Balance all atoms other than H and OOxidation : SO2 SO42-Reduction : Cr2O72- 2Cr3+Balance O by adding H2O to O deficient sideOxidation : SO2 +2H2O SO42-Reduction : Cr2O72- 2Cr3+ +7H2OBalance H by adding H+ to H deficient sideOxidation : SO2 +2H2O SO42- +4H+Reduction : Cr2O72-+14H+ 2Cr3+ +7H2OBalance charge by adding electronsOxidation : SO2 +2H2O SO42- +4H++2e-Reduction : Cr2O72-+14H+ +6e- 2Cr3+ +7H2OMultiply the oxidation half by 3 and add both equations3SO2 + Cr2O72-+2H+ 3SO42- +2Cr3+ +H2OThis is the required balanced equation.

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