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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 - Chemistry

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प्रश्न

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%.

संख्यात्मक

उत्तर

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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पाठ 9: Electro Chemistry - Evaluation [पृष्ठ ६६]

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सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 12 TN Board
पाठ 9 Electro Chemistry
Evaluation | Q 13. | पृष्ठ ६६
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