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100 mL of a 0.05 M CuSO4 aqueous solution was electrolysed using inert electrodes by passing current till the pH of the resulting solution was 2. The solution after electrolysis was neutralised and then treated with excess KI and the formed I2 was titrated with 0.04 M Na2S2O3. Calculate the volume of Na2S2O3 required in mL.

A
112.5 mL
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B
100 mL
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C
125 mL
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D
None of these
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Solution

The correct option is A 112.5 mL

During electrolysis of CuSO4
At cathode,
Cu2+ + 2e Cu(s)
Eo = 0.34

At anode,
2H2O O2 + 4H+ + 4e Eo = 1.23

Therefore some of Cu2+ is converted to Cu(s) and H+ is released at anode

pH=2 hence, [H+] = 102M
[SO24] = 1022M
=5× 103M

Initial milli moles of CuSO4 = 50
Amount of CuSO4 left after electrolysis =505
=45 milli moles
Cu2+ = 0.045M
Also,
CuSO4 + KI CuI + I2
I2 + Na2S2O3 Na2S4O6 + NaI
As Cu2+ is reacting with KI,
Therefore Meq Cu2+ = Meq KI
Also Meq KI = Meq I2
Meq I2 = Meq Na2S2O3
Meq Cu2+ = 0.045 × 1 × 100
= 4.5

Meq Na2S2O3 = 0.04 × 1 × VNa2S2O3 0.04 × 1 × VNa2S2O3 = 4.5
VNa2S2O3 = 4.50.04
=112.5mL


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