Question

Balance the reaction :- CrO4- + I- → Cr+3 + IO-3 ( Basic medium ) ​

Answer 1

Answer:

$\boxed{\red{\sf 5H_2O +CrO_4^- + I^- \longrightarrow Cr^{+3} + IO_3^-+ 10OH^- }}$

Explanation:

Given Rxⁿ :-

$\sf CrO_4^- + I^- \longrightarrow Cr^{+3} + IO_3^-$

Split the reaction :-

$\sf CrO_4^{-2}\longrightarrow Cr^{+3}\qquad\dots(i)$

• Oxidation number of Cr changes from +6 to +3 . Hence the n factor = (6-3) = 3

$\sf I^- \longrightarrow IO_3^- \qquad\dots (ii)$

• Oxidation number of I changes from -1 to +5 .Hence the n factor = [ 5-(-1)] = 6

2(i) + (ii) :-

$\sf 2CrO_4^- + I^- \longrightarrow 2Cr^{+3} + IO_3^-$

The number of oxygen atoms on LHS is 8 and that on RHS is 3 . Since the medium is basic , add 5 water molecules to LHS ,

$\sf 5H_2O+CrO_4^- + I^- \longrightarrow Cr^{+3} + IO_3^-$

Finally , add double number of $\sf OH^-$ ions on RHS ,

$\sf 5H_2O +CrO_4^- + I^- \longrightarrow Cr^{+3} + IO_3^-+ 10OH^-$

This is the required answer !

Answer 2

Explanation:

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