Estimate the mass of copper metal produced during the passage of 5 A current through `CuSO_(4)` solution for 100 minutes. The molar mass of Cu is 63.5 g `mol^(-1)`

Answer

Mass of Cu metal formed `=9 .87 g`

Explanation

Given : Electric current = I = A
Time = t = 100 min = 100 `xx 60 s = 6000 s `
Molar mass of Cu = Atomic mass of Cu = 63.5 g `mol^(-1)`
Mass of cu deposited = ?
Reduction half reaction at cathode .
`Cu_((ag))^(++) + 2e^(-) to Cu_((s))`
Quantity of electricity `= Q= I xx t = 5 xx 6000 = 3000 C`
Number of moles of electrons passed `= (Q)/(F ) `
`= (3000)/(96500) "mol"`
` C "mol"^(-1)`
`= 0.3109 "mol"`
From the half reduction reaction 1 mol of Cu is formed by the passage of 2 moles of electrons.
Molar ratio of Cu `= ("Molar of Cu formed " )/("Number of moles of electron " ) `
`= (1 " mol " ) /(2 " mol " ) =0.5 `
Number of moles of Cu formed
= Number of moles of electrons passed `xx ` Mole ratio
`= 0.3109 xx 0.5 = 0.1554` mol
Mass of Cu formed = Moles of Cu formed `xx` Molar Mass of Cu
`= 0.1554 xx 63.5 = 9.87 g`