CHEMISTRY TEXTBOOK

b. Suppose two cells containing different
electrolytes are connected in series. The same
quantity of electricity is passed through them.
The masses of the substances liberated at the
electrodes of the two cells are related as given
below :

The mass of the substance produced at
the electrode of first cell is given by

W 1 =

Q(C)

96500 (C/mol e)

× (mole ratio) 1 × M 1

Hence, Q(C)
96500 (C/mol e)

= W^1
(mole ratio) 1 × M 1
Similarly mass of substance liberated
at the electrode of second cell is W 2 in the
equation,
Q(C)
96500 (C/mol e)

=

W 2

(mole ratio) 2 × M 2

M 1 and M 2 are the molar masses of
substances produced at the electrodes of cells
1 and 2.

Because Q(C)

96500 (C/mol e)

is the same
for both,

We have

W 1

(mole ratio) 1 × M 1

= W^2
(mole ratio)^2 × M^2

(5.21)

Problem 5.6 : What is the mass of Cu

metal produced at the cathode during

the passage of 5 ampere current through

CuSO 4 solution for 100 minutes. Molar

mass of Cu is 63.5 g mol-1.

Solution :

i. Stoichiometry for the formation of Cu is

Cu^2 ⊕ (aq) + 2 e = Cu (s)

Hence,

mole ratio = 1 mol

2 mol e

ii. Mass of Cu formed,

W =

I (A) × t (s)

96500 (C/mol e)

× mole ratio × molar mass

of Cu

= 5 A × 100 × 60 s

96500 (C/mol e-)

×

1 mol

2 mol e-^

× 63.5 g mol-1

= 9.87 g

Problem 5.7 : How long will it take to

produce 2.415 g of Ag metal from its salt

solution by passing a current of 3 ampere?

Molar mass of Ag is 107.9 g mol-1.

Solution :

i. Stoichiometry :

Ag⊕ (aq) + e Ag (s)

mole ratio = 1 mol

1 mol e

ii. W =

I (A) × t (s)

96500 (C/mol e)

× mole ratio × molar mass

of Ag

2.415 g = 3 A × t

96500 (C/mol e)

×

1 mol

1 mol e^

× 107.9 g mol-1

t = 2.415 × 96500 (C = As)

3 A × 107.9

= 720 s = 12 min.

Do you know?

Names, galvanic or voltaic

are given in honour of Italian

scientists L. Galvani and A. Volta for their

work in electrochemistry.