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Question
A current of 1.608A is passed through 250 mL of 0.5 M solution of copper sulphate for 50 minutes. Calculate the strength of Cu2+ after electrolysis assuming volume to be constant and the current efficiency is 100%.
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Solution
Given I = 1.608A
t = 50 min = 50 × 60 = 3000 S
V = 250 mL
C = 0.5M
η = 100%
Calculate the number of Faradays of electricity passed through the CuSO4 solution
Q = It
Q = 1.608 × 3000
Q = 4824 C
∴ Number of Faradays of electicity = `(4824 "C")/(96500 "C")` = 0.05F
Electrolysis of CuSO4
\[\ce{Cu^{2+}_{( aq)} + 2e^- -> Cu_{(s)}}\]
The above equation shows that 2F electricity will deposit 1 mole of Cu2+ to Cu.
∴ 0.05F electricity will deposit
`(1 "mol")/(2"F") xx 0.05 "F"` = 0.025 mol
Initial number of molar of Cu2+ in 250 ml of solution
= `0.5/(1000 "mL") xx 250 "mL"`
= 0.125 mol
∴ Number of moles of Cu2+ after electrolysis
= 0.125 − 0.025
= 0.1 mol
∴ Concentration of Cu2+
= `(0.1 "mol")/(250 "mL") xx 1000 "mL"`
= 0.4 M
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