What will be the ratio nCr2O2−7nSO2−3 when the following redox reaction Cr2O2−7(aq)+SO2−3(aq)→Cr3+(aq)+SO2−4(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 B1:3 The redox reaction is given below : +6Cr2O2−7+3+4SO2−3→+3Cr3+++6SO2−4
So , oxidizing agent : Cr2O2−7 Reducing agent : SO2−3
n-factor : for Cr2O2−7:6 , for SO2−3:2 nf ratio is 1:3 Cross multiply the oxidising and reducing agent with nf ratio,
Cr2O2−7+3SO2−3→Cr3++SO2−4
Balancing all atoms except O and H :
Cr2O2−7+3SO2−3→2Cr3++3SO2−4
Balance oxygen atoms by adding H2O
Cr2O2−7+3SO2−3→2Cr3++3SO2−4+4H2O Balance hydrogen atoms by adding H+
Cr2O2−7+3SO2−3+8H+→2Cr3++3SO2−4+4H2O
Balance charge charge in reactant side = 0 charge in product side = 0
So the balanced equation is :
Cr2O2−7+3SO2−3+8H+→2Cr3++3SO2−4+4H2O So the ratio nCr2O2−7nSO2−3=13