Home » Past Questions » Chemistry » Calculate the mass of copper deposited when a current of 0.5 ampere was passed t...

Calculate the mass of copper deposited when a current of 0.5 ampere was passed t...


Question

Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]

Options

A) 0.300g

B) 0.250g

C) 0.2242g

D) 0.448g


The correct answer is D.

Explanation:

M = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)

= \(\frac{MmIT}{96500n}\)

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

Mass of compound deposited = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol-1

no of charge = + 2

Mass = \(\frac{64 \times 1350}{96500 \times 2}\)

Mass = 0.448g


More Past Questions:


Discussion (2)

  • The equation for the electrolytic reaction is;
    \( CuCl_2\rightarrow Cu^{2+} +2Cl^{-} \)
    Mass =[M]
    Molecular mass [Mm]=\( 64gmol^{-1} \)
    Current [I]= 0.5A
    Time [T] = 45mins= 45 × 60= 2700sec
    charge [n]=2
    \( M=\frac{MmIt}{96,500n} \)
    \( M= \frac{64×0.5×2700}{96500×2} \)
    \( M= \frac{86,400}{193,000} \)
    M=0.4477g
    M≈ 0.448g
    Option [D] of s very much correct.

  • M = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)

    = \(\frac{MmIT}{96500n}\)

    Copper II Chloride = CuCl2

    CuCl2 → Cu2+ + 2Cl2

    Mass of compound deposited = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)

    Q = IT

    I = 0.5A

    T = 45 × 60

    T = 2700s

    Q = 0.5 × 2700

    = 1350c

    Molarmass = 64gmol-1

    no of charge = + 2

    Mass = \(\frac{64 \times 1350}{96500 \times 2}\)

    Mass = 0.448g

    Reply
    Like