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Question

Balance the following equation by ion electron method-
Cu2O+H++NO3Cu2++NO+H2O

A
3Cu2O+14H++2NO36Cu2++2NO+7H2O
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B
3Cu2O+14H++4NO36Cu2++4NO+7H2O
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C
6Cu2O+14H++NO36Cu2++2NO+7H2O
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D
None
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Solution

The correct option is A 3Cu2O+14H++2NO36Cu2++2NO+7H2O
The unbalanced redox equation is as follows:
Cu2O+H++NO3Cu2++NO+H2O

Assign the oxidation number to each element present in the reaction.
+1Cu2O+H+++5NO3+2Cu2+++2NO+H2O

Split the reaction into two halves i.e. the oxidation half and the reduction half.
Balance the atom other than oxygen and hydrogen atom in both half reactions.
Oxidation half equation:
+1Cu2O2Cu2+
Reduction half equation:
+5NO3+2NO
Cu and N are balanced.

Now, balance O atoms by adding water molecules on the side containing less number of O atoms and balance hydrogen atoms by adding H+ ions to the side containing less number of H atoms.
Oxidation half equation- Cu2O+2H+2Cu2++H2O---(i)
Reduction half equation- 4H++NO3NO+2H2O---(ii)

Equation (i) and (ii) are balanced atom wise. Now balance both equations charge wise. To balance the charge, add electrons to the electrically positive side.
Oxidation half equation-
Cu2O+2H+2Cu2++H2O+2e
Reduction half reaction-
4H++NO3+3eNO+2H2O

Multiply the half equations by suitable factors to equalize the number of electrons in the two half equations.
Oxidation half reaction- 3Cu2O+6H+6Cu2++3H2O+6e
Reduction half reaction- 8H++2NO3+6e2NO+4H2O

Add the two half equations and cancel the number of electrons on both sides.
3Cu2O+14H++2NO36Cu2++2NO+7H2O


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