Question

I 7) Small amounts of the metal strontium (Sr, MW - 87.62 g/mol) can be produced by electrolysis. The half-reaction for the production of strontium metal is Sr+ + 2 e Sr Electrolysis is carried out using a current i - 12.0 amp (1 amp = 1 C/s). How long in minutes) will it take to produce 12.0 g of strontium metal? [12 points)

Answer 1

Answer :

Given, a constant current of 12.0amp is passed through an electrolytic cell to produce 12.0g of strontium metal.

The reduction half cell reaction is :

Sr2+ (aq) + 2e Sr(s)

Thus, here the change of the oxidation state of Strontium due to deposition is = 2

The molar weight of Sr is 87.62 g/mol

We know, in case of a redox reaction, the equivalent weight of a substance = $\frac{\text{molar mass}}{\text{change of oxidation state of the substance in the redox reaction}}$ Thus, the equivalent weight of the Strontium is = $\frac{87.62g}{2} = 43.81 g$

Thus, when a constant current is passed in the electrolytic cell, strontium metal is deposited in the cathode electrode. We have to determine the time required to deposit 12.0g of Strontium metal in the cathode electrode.

We can solve the problem by using the Faraday's 1st law of electrolysis.

Faraday's 1st law :

The amount of deposited metal (w) at cathod electrode is directly proportional to the quantity of electricity passed(Q).

i.e.

Wa a Q

Or,

W = Zo

Where, Z = electrochemical equivalent of the deposited substance.

Or,

W = Z*I*t

Where, I = current strength or constant current passed

t = time of the deposition

Now,$Z = \frac{\text{equivalent-weight}}{F}$

Thus, $w = \frac{\text{equivalent-weight}}{F} *I*t$

Thus, for the given problem :

$12.0g = \frac{43.81 g}{96500C } *12.0 amp*t$

We know, 1 amp = 1 C/sec

Or, $12.0g = \frac{43.81 g}{96500C } *12.0 C/sec*t$

Or, $t = \frac{12.0g *96500C }{43.81 g *12.0 C/sec}$

Or, $t = \frac{96500}{43.81} \text{ sec} = 2203 sec$

Thus, we need to flow a constant current of 12.0 amp for 2203 sec = 36 min 43 sec = 36.72 minutes to deposit 12.0g of Strontium metal.