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Question

Write balanced redox reactions of the following:

$${ H }_{ 2 }{ O }_{ 2 }(aq)$$ reduces $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)$$ to green coloured $${ Cr }^{ 3+ }(aq)$$ in acidic medium.


A
Cr2O27+3H2O2+8H+2Cr3++3O2+7H2O
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B
Cr2O29+3H2O2+4H+2Cr3++3O2+7H2O
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C
Cr2O27+3H2O2+8H+2Cr3++3O2+9H2O
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D
None of these
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Solution

The correct option is B $${ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+3H }_{ 2 }{ O }_{ 2 }+8{ H }^{ + }\longrightarrow 2{ Cr }^{ 3+ }+3{ O }_{ 2 }+7{ H }_{ 2 }O$$
$${ H }_{ 2 }{ O }_{ 2 }(aq)$$ reduces $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)$$ to green coloured $${ Cr }^{ 3+ }(aq)$$ in acidic medium.

The unbalanced redox equation is as follows:
$${ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+H }_{ 2 }{ O }_{ 2 }\longrightarrow { Cr }^{ 3+ }+{ O }_{ 2 }$$

Balance all atoms other than H and O.
$${ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+H }_{ 2 }{ O }_{ 2 }\longrightarrow 2{ Cr }^{ 3+ }+{ O }_{ 2 }$$

The oxidation number of Cr changes from 6 to 3. The change in the oxidation number per Cr atom is 3. For 2 Cr atoms, the change in the oxidation number is 6.

The oxidation number of O changes from -1 to 0. The change in the oxidation number per O atom is 1. For 2 O atoms, the change in the oxidation number is 2.

To balance the increase in the oxidation number with a decrease in the oxidation number multiply $$H_{ 2 }O_2$$ and $$O_{2 }$$ with 3.

$${ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+3H }_{ 2 }{ O }_{ 2 }\longrightarrow 2{ Cr }^{ 3+ }+3{ O }_{ 2 }$$

To balance O atoms, add 7 water molecules on RHS.

$${ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+3H }_{ 2 }{ O }_{ 2 }\longrightarrow 2{ Cr }^{ 3+ }+3{ O }_{ 2 }$$

To balance H atoms, add 8 $$H^+$$ on LHS.

$${ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+3H }_{ 2 }{ O }_{ 2 }+8{ H }^{ + }\longrightarrow 2{ Cr }^{ 3+ }+3{ O }_{ 2 }+7{ H }_{ 2 }O$$

This is the balanced chemical equation.

Chemistry

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