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Question

# Balance the following equations by ion electron method.a. Cr2O2−7+C2H4O+H⊕⟶2Cr3++C2H4O2+H2Ob. Cu2O+H⊕+NO⊖3⟶Cu2++NO+H2O

A
a. Cr2O27+8H+3C2H4O2Cr3++4H2O+3C2H4O2
b. 3Cu2O+14H+2NO36Cu2++7H2O+2NO
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B
a. Cr2O27+4H+3C2H4O2Cr3++2H2O+3C2H4O2
b. 3Cu2O+14H+4NO36Cu2++7H2O+4NO
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C
a. 2Cr2O27+8H+3C2H4O4Cr3++4H2O+3C2H4O2
b. 6Cu2O+14H+2NO36Cu2++7H2O+2NO
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D
None of these
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Solution

## The correct option is A a. Cr2O2−7+8H⊕+3C2H4O⟶2Cr3++4H2O+3C2H4O2 b. 3Cu2O+14H⊕+2NO⊖3⟶6Cu2++7H2O+2NO(a) The unbalanced redox equation is as follows:Cr2O2−7+C2H4O+H⊕⟶2Cr3++C2H4O2+H2OThe oxidation numbers for all the atoms are assigned. Cr26O2−7+C2−1H4O+H⊕⟶2Cr3+3+C20H4O2+H2OOxidation half equation: C2H4O→C2H4O2Reduction half equation: Cr2O2−7→2Cr3+All atoms other than H and O are balanced.Balance O atoms by adding water molecules on the side containing less number of O atoms.Balance hydrogen atoms by ading H+ ions to the side containing less number of H atoms.Oxidation half equation: C2H4O+H2O→C2H4O2+2H+Reduction half equation: Cr2O2−7+14H+→2Cr3++7H2OBalance charges by adding appropriate number of electrons.Oxidation half equation: C2H4O+H2O+2e−→C2H4O2+2H+Reduction half equation: Cr2O2−7+14H+→2Cr3++7H2O+6e−Multiply the half equations by suitable factors to equalize the number of electrons in the two half equations.Oxidation half equation: 3C2H4O+3H2O+6e−→3C2H4O2+6H+Reduction half equation: Cr2O2−7+14H+→2Cr3++7H2O+6e−Add the two half equations and cancel the number of electrons on both sides.Cr2O2−7+3C2H4O+8H⊕⟶2Cr3++3C2H4O2+4H2OThis is the balanced chemical equation.(b) The unbalanced redox equation is as follows:Cu2O+H⊕+NO⊖3⟶Cu2++NO+H2OThe oxidation numbers for all the atoms are assigned Cu2+1O+H⊕+N+5O⊖3⟶Cu2++2+NO2+H2OOxidation half equation: Cu2O→Cu2+Reduction half equation: NO⊖3→NOAll atoms other than H and O are balanced.Oxidation half equation: Cu2O→2Cu2+Reduction half equation: NO⊖3→NOBalance O atoms by adding water molecules on the side containing less number of O atoms.Balance hydrogen atoms by ading H+ ions to the side containing less number of H atoms.Oxidation half equation: Cu2O+2H+→2Cu2++H2OReduction half equation: NO⊖3+4H+→NO+2H2OBalance charges by adding appropriate number of electrons.Oxidation half equation: Cu2O+2H+→2Cu2++H2O+2e−Reduction half equation: NO⊖3+4H++3e−→NO+2H2OMultiply the half equations by suitable factors to equalize the number of electrons in the two half equations.Oxidation half equation: 3Cu2O+6H+→6Cu2++3H2O+6e−Reduction half equation: 2NO⊖3+8H++6e−→2NO+4H2OAdd the two half equations and cancel the number of electrons on both sides.3Cu2O+14H⊕+2NO⊖3⟶6Cu2++7H2O+2NOThis is the balanced chemical equation.

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