Question

An alloy of lead (valency = 2) and thallium (valency = 1) containing 65% Pb and 35% Tl, by weight, can be electroplated onto a cathode from a solution. How many grams of this alloy will deposit in 9.65 hours using a constant current of 0.5 amp? [Given that only these two elements deposit simultaneously in given mass ratio at cathode]
[Atomic weight Pb = 208, Tl = 200]
 (A) 3.7 gm
 (B) 22.5 gm
 (C) 37 gm
 (D) 24.9 gm

Answer 1

Hint: First find out the individual amounts of lead and thallium that should be deposited on the electrodes using the Faraday’s Law and formula: $m = \dfrac{{I.t.Z}}{F}$. Now find the weight of 65% of Pb and 35% of Tl from it. Then the total weight of the alloy deposited will be the sum of weight of Pb and Tl.

Formula used:
-Faraday’s Law: $m = \dfrac{{I.t.Z}}{F}$ (1)
Where, m = weight deposited at electrode;
               I = current passing through it
               t = time
               F = Faraday’s constant = 96500,
               Z = Electrochemical equivalent of substance deposited.


Complete step by step answer:
-To find out the amount of alloy deposited we will be using the Faraday’s Law. So, first let us see what the law says.
According to this law: The mass of a substance produced or deposited at an electrode during electrolysis is proportional to the number of moles of electrons or the quantity of electricity transferred to the electrode.
Mathematically it can be written as: $m = \dfrac{{q.Z}}{F}$ and since q = It, we can also write it as:
                                    $m = \dfrac{{I.t.Z}}{F}$ (1)

-First we will calculate the mass of Pb that should be deposited under given conditions using equation (1).The question has given us: valency of Pb = 2, t = 9.65 hours (9.65 × 60 × 60 seconds) and current = 0.5 amp.

The value of Z for Pb = $\dfrac{{At.wt.}}{{valency}}$ = $\dfrac{{208}}{2}$ = 104
Now putting these values in equation (1):
                            $m = \dfrac{{I.t.Z}}{F}$
                                   = $\dfrac{{0.5 \times (9.65 \times 60 \times 60) \times 104}}{{96500}}$
                                   = 18.72 gm

But since the question says that only 65% of Pb was deposited. We will find out the weight of 65% of Pb that should have been deposited.
65% of m = 65% of 18.72 = $\dfrac{{65}}{{100}} \times 18.72$
                                             = 12.168 gm of Pb

-Now we will find out the weight of Tl that should have been deposited under given conditions. The valency of Tl is given to be =1, t = 9.65 hours (9.65 × 60 × 60 seconds) and current = 0.5 amp.
The value of Z for Tl = $\dfrac{{At.wt.}}{{valency}}$ = $\dfrac{{200}}{1}$ = 200
Now putting these values in equation (1):
                              $m = \dfrac{{I.t.Z}}{F}$
                                     = $\dfrac{{0.5 \times (9.65 \times 60 \times 60) \times 200}}{{96500}}$
                                     = 36 gm

But since the question says that only 35% of Tl was deposited. We will find out the weight of 35% of Tl that should have been deposited.
35% of Tl = 35% of 36 = $\dfrac{{35}}{{100}} \times 36$
                                       = 12.6 gm of Tl

-Now we can find out the total weight of alloy that was deposited by adding the weights of Pb and Tl that was deposited.
Total weight of alloy = weight of Pb + weight of Tl
                                     = 12.168 + 12.6
                                     = 24.768 gm ≈ 24.9 gm
Hence the total amount of alloy deposited will be 24.9 gm.
So, the correct answer is “Option D”.

Note: Remember that the weight (m) we obtain from the Faraday’s Law are the weights that should have been deposited and not the weights that are deposited. So, for the weights that are deposited we multiply ‘m’ by their respective percentages.
Also the electrochemical equivalent of a substance is basically the mass of the substance deposited when a current of 1 amp is passed for 1 second. So, it can also be said to be equivalent weight.