Question

How many grams of copper will be deposited at cathode when an aqueous solution of copper sulphate is electrolysed with a current of 6 amperes for 965 seconds? (Atomic mass of copper = 63.5 u)
2.8 g
3.89
5.2g
1.99 1​

Answer 1

Given:

Solution of the Copper sulphate is electrolyse with having current of 6 ampere for  a time of 965 seconds and atomic no. of copper is 63.5 u

To Find:

Grams of copper will become deposited at cathode electrode when aq solution of the copper sulphate electrolysed = ?

Solution:

We know Mass deposited during electrolysis formula

$W = \dfrac{Mit}{F z}$

M = 63.5 $gm$

Current = $i$ = 6 Ampere

Time = t  = 965 seconds

Atomic mass = M = 63.5

$W =$  6× 965 × 63.5 / 96500 × 2

W = 1.905  $gm$

Answer 2

Given:

Current applied to electrolysed the aqueous solution of copper sulphate        = 6 Ampere

Time taken to electrolysed the aqueous solution of copper sulphate                        = 965 seconds

To find:

Mass of copper in grams deposited at cathode.

Solution:

Firstly find the total charge flowing through the aqueous solution of copper sulphate having current 6 Ampere for time 965 seconds.

So, formula used to find total charge is-

              Q = I × t

here, Q = total charge

          I = current = 6 Ampere = 6 Cs⁻¹

          t = time taken = 965 seconds

by putting all the values, we get

        Q = 6 Cs⁻¹ × 965 s

         Q = 5790 C

So, total charge is 5790 Coulombs.

Now, find the total of moles of copper that will deposited at cathode, formula used will be-

           $n = \frac{Q}{zF}$

here, n = number of moles

        Q = total charge = 5790 C

        z = number of electrons involved in the half-cell = 2 electrons

        F = Faraday constant = 96500 Cmol⁻¹

by putting all the values, number of moles will be,

           $n = \frac{5790 C}{2 \times 96500 Cmol^-^1}$

           n = 0.03 mol

So, the total number of moles of copper deposited is 0.03 mol.

therefore, mass of copper deposited at cathode will be,

                   $n = \frac{w}{MM}$

or,               w = n × MM

here, w = mass of copper deposited at cathode

         n = number of moles of copper deposited at cathode = 0.03 mol

        MM = atomic mass of copper = 63.5 u

by substituting all the values mass of copper will be calculated as,

        w = 0.03 mol × 63.5 gmol⁻¹

        w = 1.905 g

So, mass of copper deposited at cathode is 1.9g

Therefore, option(d) is correct.