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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry > Unit 14

Lesson 6: Electrolytic cells and electrolysis# Quantitative electrolysis

Calculating how much zinc deposits on the zinc electrode after 1.0 h when a current of 5.0 A is applied to the battery. Created by Jay.

## Want to join the conversation?

- Can anybody clarify this for me?

In electrolytic cell :-

ANODE - is the source of electron --> hence losses e- and becomes oxidation.

CATHODE - gains e- (from external battery) and becomes a reduction.

Thank you in advance !!(6 votes)- Yes correct. You can also think of OIL/RIG of electrons : Oxidation Is Loss / Reduction Is Gain(11 votes)

- How did you get to 96,500?

charge on single electron = 1.6 * 10^-19 C (e)

charge on one mole electron = e * N.A

= e* 6.022 * 10^23 = 96352 ??(4 votes)- 6.0221 × 10²³ electrons × 1.6022 × 10⁻¹⁹ C/electron = 96 486 C.

He used a rounded-off number.(5 votes)

- what if the moles of Cu 2+ formed were asked for the same data, would they be the same as that of Zn?(5 votes)
- In this equation one mole of oxidised Cu (from solid to liquid) is needed to reduce one mole of Zn (from liquid to solid). So the molarities are the same however because Cu has slightly different molar mass than Zn the solid Cu bar would be only 6,0 g lighter.(2 votes)

- why is anode attract electrons even tho it positive(0 votes)
- i know its useless to reply to your comment 4 years later, but isnt it the basics to know that the anode always attracts negatively charged ions (anions) and the cathode attracts positive ions (cations)(2 votes)

- Why is electrochemical series not according chemical reactivity in periodic table?(2 votes)
- What's the point of an electrolytic cell? Isn't it just a waste cuz you're using a battery to make a battery? Thanks.(2 votes)
- In an electrolytic cell, isn't there only one cell and no salt bridge?(1 vote)
- Can't we solve this question in this manner:

Q=5×60×60=18000 Coulumbs

Then,

2 mol of e- gives 1 mol of Zn+2 Zn=65g

So,

2×96,500=193000

1800/193000 × 65g=6.02g deposited(Answer)(1 vote) - Why do we need znso4 solution in galvanic cell in the zinc electrode container and why not anything else like mgso4 as we only need so4- ions to combine it with zn2+ ions?(1 vote)
- Couldn't you say that 1 A is = 6.24*10^18 passed per second to calculate how many electrons would move in an hour, and then work out how many Zn2+ ions would be reduced from that?(1 vote)
- 6.24*10^18 what per second?(1 vote)

## Video transcript

- [Voiceover] Here's
the electrolytic cell we talked about in the previous video. Remember, an electrolytic
cell uses current to drive a non-spontaneous redox reaction. So we need an external voltage source and so here's our battery. We know electrons come
from the negative terminal of the battery and electrons are forced onto the zinc electrode. We know there are zinc
two plus ions in solution so if zinc two plus gains two electrons zinc two plus is reduced to solid zinc, so we're going to form solid zinc at our zinc electrode so
solid zinc forms here. At our other electrode,
the battery pulls electrons away from copper and
solid copper is oxidized so we lose two electrons
to form Cu two plus, so the copper electrode
loses mass over time. When we look at our problem, this is a quantitative
electrolysis problem because they're telling
us what the current is, 5.0 amps, so we're applying
a current of 5.0 amps. How much zinc, in grams, deposits on the zinc electrode after one hour? So we have to figure out how much zinc forms on our zinc electrode. So first we need to think about
the definition for current, so current, let me write this down here, current is equal to charge over time. So in physics, I is equal to current, charge is represented by Q,
and time is lower-case t. We have already seen charge
before in earlier videos and we know that's measured in Coulombs and time is in seconds,
so Coulombs per second gives us an ampere, or an amp for short. So let's plug in what we know. We know the current is five amps so we plug that into
here so we get 5.0 amps. We don't know what the charge
is but we do know the time. Alright, so how much time
are we talking about here? One hour, we need to convert
one hour into seconds so how many seconds is one hour? There are 60 minutes in
an hour and each minute is 60 seconds so 60
times 60 gives us 3,600 so there are 3,600 seconds in one hour. Now we can solve for Q, we
can solve for the charge. So five times 3,600 is equal to 18,000. So after one hour, we're
talking about 18,000 Coulombs. >From the charge, we
can figure out how many moles of electrons we're dealing with here because of Faraday's constant. So remember, Faraday's constant tells us that one mole of electrons has
a charge of 96,500 Coulombs. So if we have 18,000 Coulombs
and we're trying to find how many moles of electrons that is, we would need to divide
by Faraday's constant. So 18,000 divided by 96,500,
which is Faraday's constant, the charge of one mole of electrons. If you do it this way,
you can see that Coulombs would cancel out and you
would get moles of electrons. So let's do that on the calculator. 18,000 divided by 96,500 gives us 0.19 so this is equal to
0.19 moles of electrons. So 0.19 moles of electrons were forced through our electrolytic
cell because of the battery. So next we need to relate
the moles of electrons to the moles of zinc that are formed and we can do that by remembering our reduction half reaction. Zinc two plus plus two
electrons forms solid zinc so let's write down our
reduction half reaction here. So we know that zinc two plus ions are being reduced to form solid zinc. So let's think about
those mole ratios here. Two moles of electrons
are needed to reduce one mole of zinc two plus ions to form one mole of solid zinc. So we now have the relationship. Alright, we know the
mole ratio of electrons to moles of solid zinc, it's a mole ratio of two to one. So one mole of zinc is produced for every two moles of
electrons that we have. So let's set up a proportion to figure out how many moles of zinc are produced. So we'll put electrons over solid zinc, so we have a mole ratio of two to one, so our mole ratio is two to one. And then on the right
side of our proportion, alright, we know that
0.19 moles of electrons were forced through our cell so we can write down here 0.19 moles of electrons and that will be over x and x represents the moles of solid zinc. Alright, so let's solve our proportion. That's two x is equal to 0.19, so two x is equal to 0.19, so 0.19 divided by two is equal to 0.095, so x is equal to 0.095, and x represents the moles. Alright, this is the moles of zinc that are produced in our reaction and again we got this from our mole ratio. One mole of zinc forms for
every two moles of electrons, so if you have 0.19 moles of electrons, half that is how many moles
of zinc that are formed? Finally, we need to go from moles of zinc to grams of zinc, alright,
our problem asks us for grams of zinc that were deposited. So going from moles to
grams is pretty easy, you just multiply by the molar mass. So if you have 0.095 moles of zinc and we multiply by the molar
mass of zinc, which is 65.39, so the molar mass of zinc
is 65.39 grams per mole. If you multiply, the moles cancel out and that gives you grams,
so let's do that math. So we have 0.095 times 65.39 which is equal to 6.2 so we get 6.2 grams of zinc. That's our final answer,
that's how much zinc deposits on the zinc electrode. So there are other ways to think about these types of problems. I prefer to go through
everything step by step and logically come to the final answer. You could also just think about
units and do all this math. So that's another way to do it. So this is just the way that I prefer to do a quantitative electrolysis problem.