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Question

What will be the ratio nCr2O27nSO23 when the following redox reaction
Cr2O27 (aq)+SO23 (aq)Cr3+ (aq)+SO24 (aq)
is balanced in an acidic medium. Where n represents the stoichiometric coefficient of the respective species.

A
2:1
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B
1:3
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C
1:2
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D
2:3
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Solution

The correct option is B 1:3
The redox reaction is given below :
+6Cr2O27+3+4SO23+3Cr3+++6SO24

So , oxidizing agent : Cr2O27
Reducing agent : SO23

n-factor :
for Cr2O27:6 ,
for SO23:2
nf ratio is 1:3
Cross multiply the oxidising and reducing agent with nf ratio,

Cr2O27+3SO23Cr3++SO24

Balancing all atoms except O and H :

Cr2O27+3SO232Cr3++3SO24

Balance oxygen atoms by adding H2O

Cr2O27+3SO232Cr3++3SO24+4H2O
Balance hydrogen atoms by adding H+

Cr2O27+3SO23+8H+2Cr3++3SO24+4H2O

Balance charge
charge in reactant side = 0
charge in product side = 0

So the balanced equation is :

Cr2O27+3SO23+8H+2Cr3++3SO24+4H2O
So the ratio nCr2O27nSO23=13

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