Question

A current of 1.7 amperes is passed through 300 ml of 0.16 M solution of ZnS04 for 230 secs with a current efficiency of 90 percent. find out the molarity of Zn 2+ after deposition of Zn. assume volume of solution to remain constant during electrolysis.

Answer 1

let me tell how i did it @Vincent-Lyon.Fr and tell me why it is not correct -

Answer 2

Zn 2+ + 2e- ----> Zn;
0.16 M = no of moles/ 0.3 litres;
no of moles of Zn 2+ present = 0.048 moles;
=0.048 X 65.4/2 = 1.56 g. (65.4/2 = eq. wt of Zn 2+)
mass of Zn deposited =
32.7 X 351.9 /96500 [ M = ECT/96500]
=0.119 g.
therefore total mass of Zn 2+ left in solution = 1.56 - 0.119 = 1.446 gms
and hence molarity = 1.446 / 0.3 = 4.82 M which is wrong as per my book.
the answer says 0.154 M

Answer 3

Zn 2+ + 2e- ----> Zn; 0.16 M = no of moles/ 0.3 litres; no of moles of Zn 2+ present = 0.048 moles = 48 mmol
ok

=0.048 X 65.4/2 = 1.56 g. (65.4/2 = eq. wt of Zn 2+) mass of Zn deposited = 32.7 X 351.9 /96500 [ M = ECT/96500] =0.119 g. therefore total mass of Zn 2+ left in solution = 1.56 - 0.119 = 1.446 gms
not necessary; uses moles or millimoles instead.

How many mmol of electrons have circulated?
How many mmol of Zn2+ have been used up?
How many mmol of Zn2+ remain in solution?

Answer 4

okay but is there any way you can get the answer through my way?..why is it wrong?

Answer 5

It is not wrong, it is just a waste of time. Because, at one point you multiply by atomic weight, and later you divide by atomic weight.

Or rather, you should have divided by atomic weight.
Now I realise what you did in the end :
quote : and hence molarity = 1.446 / 0.3 = 4.82 \(\color{red}{\text{g/L}}\)

Answer 6

Anyway, initial mass of \(Zn^{2+}\) is 3.14 g (you should not divide by 2)
And what about the 90% ?

Answer 7

yeah i used the 90 percent to calculate the current actually passed ..ideally, it would be 391 Amperes but because of 90 percent efficiency, it is around 351 amperes. with that and faraday's first law, i calculated mass of Zn deposited. and why should we not divide by 2?

Answer 8

hey will be back in a while..moms forcing me to have dinner..:(

Answer 9

Enjoy your meal!

Answer 10

yeah thanks! anyway did you figure it out?

Answer 11

Try doing it using moles, then you will understand where you went wrong.

Answer 12

okay..

Answer 13

@Vincent-Lyon.Fr - how do i do it using moles?

Answer 14

How many mmol of electrons have circulated?
How many mmol of Zn2+ have been used up?
How many mmol of Zn2+ remain in solution?

Answer 15

2 moles of electrons?

Answer 16

sorry 0.096 moles of electrons?

Answer 17

1.7 A x 230 s = 391 C

391 C / 96500 = 0.004 moles of electrons.

90% of which have been efficient for the reaction, so 0.00365 moles of e- used in the reaction :
Zn2+ + 2e- ----> Zn

Do you follow?

Answer 18

okay..

Answer 19

Now \(\Large \frac {0.00365}{2} \large =0.001825\) mole of Zn2+ was used up.

Answer 20

but this idea will never strike my mind in exams...i will have to bank on how to do i my way..:(

Answer 21

Sorry, fuse went off. I had to find it and restart the computer.

Answer 22

Remaining Zn2+ is 0.16x0.300 - 0.001825 = 0.046 mol

Accounting for concentration 0.0046 / 0.300 = 0.153 M

Answer 23

yeah i got it....but i want to know how to do it in the way i told...

Answer 24

but this idea will never strike my mind in exams...i will have to bank on how to do i my way..:(

It is not an 'idea'. Basic chemistry is finding how many moles of this react with how many moles of that. So it is logical to think in terms of amount of substances appearing or dissappearing.

Answer 25

Ok, you mean you want to use equivalent weight ?

Answer 26

yeah

Answer 27

Be patient, I am not familiar with eq weight, so I have to think hard ;-)
It just makes things more complicated, but I will try. Back in a minute.

Answer 28

okay thanks..:D

Answer 29

Ok, here it is. Some of the lines, I copied straight from your answers.

Amount of zinc present at beginning: 0.16 M x 0.3 litres = 0.048 mole
Mass of zinc in solution at beginning: 0.048 mole x 65.4 g/mol = 3.14 g

eq. wt of Zn 2+ = 65.4/2 =32.7 g/mol (e-)
mass of Zn deposited = 32.7 X 351.9 /96500 =0.119 g.

therefore total mass of Zn 2+ left in solution = \(\color {red} {3.14}\) - 0.119 = \(\color {red} {3.02}\) g
amount of zinc left in solution 3.02/65.4 = 0.0462 mol

and hence molarity = \(\color {red} {0.0462}\) / 0.3 = \(\color {red} {0.154}\) M

Answer 30

great! i realised the mistake i was doing..i was dividing mass/volume to get molarity wherein it should be moles/volume

Answer 31

Great! I have to go now, will not be back until Thursday. See you!

Answer 32

whoa! thats a long time without your help! but anyway have a nice time and thanks !

Answer 33

Welcome \(\LARGE \ddot \smile \)