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Question

Balance the following redox reactions in basc medium :

    $$Mn{ O }_{ 4 }^{ - }+{ I }^{ - }\longrightarrow Mn{ O }_{ 2 }+I{ O }_{ 3 }^{ - }$$


A
2MnO4+I+H2O2MnO2+IO3+2OH
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B
3MnO4+I+H2O4MnO2+IO3+2OH
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C
3MnO4+I+H2O5MnO2+IO3+2OH
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D
None of these
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Solution

The correct option is A $$2Mn{ O }_{ 4 }^{ - }+{ I }^{ - }+{ H }_{ 2 }O\longrightarrow 2Mn{ O }_{ 2 }+I{ O }_{ 3 }^{ - }+2{ OH }^{ - }$$
The unbalanced redox equation is as follows:

$$Mn{ O }_{ 4 }^{ - }+{ I }^{ - }\longrightarrow Mn{ O }_{ 2 }+I{ O }_{ 3 }^{ - }\quad (basic)$$
All atoms other than H and O are balanced.
The oxidation number of Mn changes from 7 to 4. The change in the oxidation number is 3.
The oxidation number of I changes from -1 to 5. The change in the oxidation number  is 6.
To balance the increase in the oxidation number with decrease in the oxidation number multiply $$Mn{ O }_{ 4 }^{ - }+{ I }^{ - }$$ and $$Mn{ O }_{ 2 }$$ with 2.
$$2Mn{ O }_{ 4 }^{ - }+{ I }^{ - }\longrightarrow 2Mn{ O }_{ 2 }+I{ O }_{ 3 }^{ - }\quad (basic)$$
To balance O atoms, add one water molecule on RHS.
$$2Mn{ O }_{ 4 }^{ - }+{ I }^{ - }\longrightarrow 2Mn{ O }_{ 2 }+I{ O }_{ 3 }^{ - } + H_2O\quad (basic)$$
To balance H atoms, add 2$$H^+$$ on LHS.
$$2Mn{ O }_{ 4 }^{ - }+{ I }^{ - }+2H^+ \longrightarrow 2Mn{ O }_{ 2 }+I{ O }_{ 3 }^{ - } + H_2O\quad (basic)$$
Since the reaction occurs in basic medium, add 2 $$OH^-$$ ions on either side.
$$2Mn{ O }_{ 4 }^{ - }+{ I }^{ - }+2H^++2OH^- \longrightarrow 2Mn{ O }_{ 2 }+I{ O }_{ 3 }^{ - } +2OH^-+ H_2O\quad (basic)$$
On LHS, 2 $$H^+$$ ions combine with 2 $$OH^-$$ ions to form 2 water molecules out of which one water molecule cancels with one water molecule on RHS.
$$2Mn{ O }_{ 4 }^{ - }+{ I }^{ - }+{ H }_{ 2 }O\longrightarrow 2Mn{ O }_{ 2 }+I{ O }_{ 3 }^{ - }+2{ OH }^{ - }$$
This is the balanced chemical equation.

Chemistry

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