Chemguide: Support for CIE A level Chemistry

This statement is about simple electrolysis calculations.

Before you go on, you should find and read the statement in your copy of the syllabus.

Electrolysis calculations are no more difficult than any other calculation from an equation. In fact, you may well have done them as a part of whatever course you did before you started doing A level.

We will just look at three examples.

Example 1

Calculate the mass of silver deposited at the cathode during the electrolysis of silver nitrate solution if you use a current of 0.10 amps for 10 minutes.

(Use the Data Booklet to find the value of the Faraday constant and the relative atomic mass of silver.)

You will find a copy of the Data Booklet towards the end of the syllabus.

The Data Booklet gives the value F = 9.65 x 10⁴ C mol^-1 (or 96500 C mol^-1 if you prefer). A_r of Ag = 108.

The first thing to do is to work out how many coulombs of electricity flowed during the electrolysis.

Number of coulombs = current in amps x time in seconds

Number of coulombs = 0.10 x 10 x 60 = 60

Now look at the equation for the reaction at the cathode:

Just as with any other calculation from an equation, write down the essential bits in words:

1 mol of electrons gives 1 mol of silver, Ag.

Now put the numbers in. 1 mol of electrons is 1 faraday.

96500 coulombs give 108 g of silver.

So, if 96500 coulombs give 108 g of silver, all you have to do is to work out what mass of silver would be produced by 60 coulombs.

Mass of silver = 60/96500 x 108 g = 0.067 g

If 96500 coulombs give 108 g, then 1 coulomb would give 108 divided by 96500 g. 60 coulombs would produce 60 times this amount.

Example 2

This example shows you how to do the calculation if the product you are interested in is a gas.

Calculate the volume of hydrogen produced (measured at room temperature and pressure - rtp) during the electrolysis of dilute sulphuric acid if you use a current of 1.0 amp for 15 minutes.

(Use the Data Booklet to find the value of the Faraday constant and the molar volume of a gas at room temperature and pressure.)

The Data Booklet gives the value F = 9.65 x 10⁴ C mol^-1 (or 96500 C mol^-1). The molar volume of a gas at rtp = 24 dm³ mol^-1.

Start by working out how many coulombs of electricity flowed during the electrolysis.

Number of coulombs = current in amps x time in seconds

Number of coulombs = 1.0 x 15 x 60 = 900

Now look at the equation for the reaction at the cathode:

Write down the essential bits in words:

2 mol of electrons give 1 mol of hydrogen, H₂.

Now put the numbers in. Two moles of electrons is 2 faradays.

2 x 96500 coulombs give 24 dm³ H₂ at rtp.

So, if 2 x 96500 coulombs give 24 dm³ H₂, work out what volume of hydrogen would be produced by 900 coulombs.

Volume of hydrogen = 900/(2 x 96500) x 24 dm³ = 0.11 dm³

Don't quote your answer beyond 2 decimal places. The current and the molar volume are only quoted to that degree of accuracy.

Example 3

This example shows you what to do if the question is reversed.

How long would it take to deposit 0.635 g of copper at the cathode during the electrolysis of copper(II) sulphate solution if you use a current of 0.200 amp.

(Use the Data Booklet to find the value of the Faraday constant and relative atomic mass of copper.)

The Data Booklet gives the value F = 9.65 x 10⁴ C mol^-1 (or 96500 C mol^-1). A_r of Cu = 63.5.

This time you can't start by working out the number of coulombs, because you don't know the time. As with any other calculation, just start from what you know most about. In this case, that's the copper, so start with the electrode equation.

Write down the important bits of this in words:

2 mol of electrons give 1 mol of copper, Cu.

Now put the numbers in. 1 mol of electrons is 1 faraday.

2 x 96500 coulombs give 63.5 g of copper.

You need to work out how many coulombs give 0.635 g of copper.

Number of coulombs = 0.635/ 63.5 x 2 x 96500 = 1930

Now what?

You know how many coulombs you need, and you know what the current was in amps. You have got all the information you need to work out the time.

Number of coulombs = current in amps x time in seconds

1930 = 0.200 x t

t = 1930/0.200 = 9650 seconds.

Don't waste time trying to convert that into minutes or hours (unless the exam question specifically asks you to).

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