What current will deposit 3.25g of zinc in 2hrs?
[Zn = 65, F = 96500 mol^-]

humphrizy

7 years ago

From the equation of the reduction of zinc ion to zinc metal in question 2 above, we can see that:
65g of zinc = 2F = 2 x 96500C (recall 1F = 96500C)
If, 65g of Zn are deposited by 2 x 96500C of electricity
3.25g of Zn will be deposited by xC of electricity
x = 3.25 x 193000/65
= 9650C
but,
Quantity of electricity (C) = Current (A) x Time (s)
9650C = I x 2 x 3600s (1hr = 3600s)
I = 9650/7200
= 1.34A

Babliwa

3 years ago

THIS IS THE SIMPLEST FORMULA

m/M =IT/cF

Where m = amount in gram = 3.25

M = molar mass of Zn = 6 5

I= current

T= TIME

c= charge = 2 for Zn

F= Faraday = 96500

From the formula above, just make "I" which is the current the subject of the formula

The the formula will be mcF/MT=I

Then fixing the figures we have

3.25*2*96500 /65*2*60*60====1.34A

Hope you like understand the whole process

Stelb7

4 years ago

M=zit
3.25*96500=65*3600*I
I=96500*3.25/65*3600
I=1.34😌

Remigiusvictory

2 years ago

Using the formula
m=MQ/nf
remember that quantity of electricity is=current*time rate of flow of the charge Q=it
where m is the mass discharged during electrolysis
Q is the quantity of electricity
n is the ionic charge of the element discharged
f is faradays constant 96500
M is the molar mass of the compound
using m=MQ/nf
mass deposited=3.25
time=60*60*2 =7200 note convert tome to seconds
molar mass of zinc=65
n=2
f=96500
3.25=65*i*7200/96500*2
3.25=468000i/193000
Cross multiply i =3.25*193000/468000
i=627250/468000=1.340A

Anela

1 year ago

m/mm = it/ nF
m= 3.25
mm= 65
i=?
t=2hrs= 60x60x2 = 7200
nF= 2×96500=193000(Zn is divalent)

3.25/65 = 7200i/193000
CROSS MULTIPLY ✖️

=193000×3.25 = 7200i×65
=627250 = 468000i
DBS BY 468000
i= 627250/468000
= 1.340A(ANS)

Teddy45

2 months ago

sorry I=mcf/MT
3.25x2x96500/65x7200
=1.34A

Teddy45

2 months ago

we can also use I=
MT
mcf