    Question

# 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 mLDuring 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+] = 10−2M [SO2−4] = 10−22M =5× 10−3M Initial milli moles of CuSO4 = 50 Amount of CuSO4 left after electrolysis =50−5 =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  Suggest Corrections  0      Similar questions  Explore more