Question

1. Electrolytic reactions, like other chemical reactions, are not 100% efficient. In a copper purification apparatus depositing Cu from a CuSO4 solution, operation for 5.47 hours at a constant current of 5.45 A deposits 29.2 g of Cu metal. What is the efficiency?

b) How long would it take to electroplate all the Zn²⁺ in 0.280 L of 0.210 M ZnSO₄ solution with a current of 2.35 A?

please answer both A and B

Answer 1

1. (a).Ans :-

Given actual yield = 29.2 g

From first law of electrolysis :

w = MIt/(96500 x valency factor)

Where, w = Weight of the substance deposited = ?

I = Current in Ampere = 5.45 A

t = Time period in second = 5.47 hours = 19692 s

Valency factor of CuSO₄ = 2

Therefore,

w = (63.5 g).(5.45 A).(19692 s) / (96500 C x 2)

= 35.31 g = Theoretical yield

So,

Efficiency = Actual yield x 100 / Theoretical yield

= 29.2 g x 100 / 35.31 g

= 82.7 %

Therefore, Efficiency = 82.7 %

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(b).Number of moles of ZnSO₄ (n) = Molarity x Volume in L = 0.210 M x 0.280 L = 0.0588 mol

From first law of electrolysis :

w = MIt/(96500 x valency factor)

w/M = It/(96500 x valency factor)

n = It/(96500 x valency factor) .....................(1)

because, number of moles (n) = w (Given mass)/ (Gram molar mass) M

I = Current in Ampere = 2.35 A

Valency factor of ZnSO₄ = 2

From equation (1) :

n = It/(96500 C x valency factor)

t = (96500 C x valency factor x n) / I

t = (96500 C x 2 x 0.0588 mol) / 2.35 A

t = 4829.106 s = 1.34 hour

Therefore, time period required = t = 1.34 hour