Classwork Series and Exercises {Chemistry- SS2}: Calculations Based on Faraday’s Law of Electrolysis

Introduction

Three equations relate these quantities:

• amperes x time = Coulombs
• 96,500 coulombs = 1 Faraday
• 1 Faraday = 1 mole of electrons

Calculating the Quantity of Substance Produced or Consumed

To determine the quantity of substance either produced or consumed during electrolysis given the time a known current flowed:

• Write the balanced half-reactions involved.
• Calculate the number of moles of electrons that were transferred.
• Calculate the number of moles of substance that was produced/consumed at the electrode.
• Convert the moles of substance to desired units of measure.

Example 1:   A 40.0 amp current flowed through molten iron (III) chloride for 10.0 hours (36,000 s). Determine the mass of iron and the volume of chlorine gas (measured at 25^oC and 1 atm) that is produced during this time.

Solution:

• Write the half-reactions that take place at the anode and at the cathode.

Anode (oxidation):  2 Cl^–  Cl_2(g) + 2 e^–

Cathode (reduction):  Fe³⁺ + 3 e^–   Fe_(s)

• Calculate the number of moles of electrons.

I = 40A, t = 10hrs (10 x 60 x 60) = 36000s

Q = I x t

       40 x 36000 = 1440000C

1 mole of Fe liberate 3F (3 x 96500)

What mole of Fe will liberate 1440000C

= 1 x 1440000 ÷ (3x 96500)

= 4.97 mole of Fe

1 mole of Cl liberate 2F (2 x 96500)

What mole of Cl will liberate 1440000C?

= 1 x 1440000 ÷ (2 x 96500)

= 7.46 mole of Cl

• Calculate the mass of iron using the molar mass and calculate the volume of chlorine gas using the ideal gas law (PV = nRT).

Molarmass of Fe = 56gmol^-1, calculated mole of Fe = 4.97

Mass of Fe deposited = mole x molarmass

                                        4.97 x 56 = 278g

Using ideal gas law: PV = nRT

V = nRT/P

V = 7.46 x 0.0821 x 298/ 1

Volume of Cl₂ = 182L

Calculating the Time Required 

To determine the quantity of time required to produce a known quantity of a substance given the amount of current that flowed:

• Find the quantity of substance produced/consumed in moles.
• Write the balanced half-reaction involved.
• Calculate the number of moles of electrons required.
• Convert the moles of electrons into coulombs.
• Calculate the time required.

Example 2:  How long must a 20.0 amp current flow through a solution of ZnSO₄ in order to produce 25.00 g of Zn metal.

Solution:

• Convert the mass of Zn produced into moles using the molar mass of Zn.

Mass of Zn = 25.00g, molarmass of Zn = 65gmol^-1

Mole = mass/molarmass

Mole = 25/65 = 0.3846mol

• Write the half-reaction for the production of Zn at the cathode.

Zn²⁺_(aq) + 2e^–  Zn_(s)

• Calculate the moles of e^– required to produce the moles of Zn and convert the moles of electrons into coulombs of charge using Faraday’s constant.

1 mole of Zn liberate 2F (2 x 96500)C

0.386 mole will liberate: (2 x 96500 x 0.3846/1) = 74231C

• Calculate the time using the current and the coulombs of charge.

Q = I x t

t = Q / I

t = 74231/20

t = 3712secs = 62 minutes = 1.03hr

Calculating the Current Required

To determine the amount of current necessary to produce a known quantity of substance in a given amount of time:

• Find the quantity of substance produced/or consumed in moles.
• Write the equation for the half-reaction taking place.
• Calculate the number of moles of electrons required.
• Convert the moles of electrons into coulombs of charge.
• Calculate the current required.

Example 3:  What current is required to produce 400.0 L of hydrogen gas, measured at STP, from the electrolysis of water in 1 hour (3600s)?

Solution:

• Calculate the number of moles of H₂. (Remember, at STP, 1 mole of any gas occupies 22.4 L.)

Moles of H₂ = 400.0/22.4

                      = 17.9mol

• The equation for this half-reaction is: 4e^– + 4H₂O_(l)  2H_2(g) + 4OH^–_(aq)
• Calculate the number of moles of electrons and convert the moles of electrons into coulombs of charge.

I mole of H₂ liberate 2F (2 x 96500)C

17.9 mole will liberate: 2 x 96500 x 17.9 = 3454700C

• Calculate the current required.

I = Q / t

  = 3454700/3600

I = 960A

Example 4: Exactly 0.4 faraday electric charge is passed through three electrolytic cells in series, first containing AgNO3, second CuSO4 and third FeCl3 solution. How many gram of rach metal will be deposited assuming only cathodic reaction in each cell?

Solution: The cathodic reactions in the cells are respectively.

Ag⁺ + e^– –> Ag

1 mole ——> 1 mole

108 g  ——-> 1 F

Cu²⁺ + 2e^– –> Cu

1 mole ——> 2 mole

63.5 g ——–> 2 F

and Fe³⁺ + 3e^– –> Fe

1 mole ——-> 3 mole

56 g ——->3 F

Hence, Ag deposited = 108 × 0.4 = 43.2 g

Cu deposited = 63.5/2×0.4 = 12.7 g

and Fe deposited = 56/3×0.4 = 7.47 g

Example 5: How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal surface of 80 cm2 with 0.005 cm thick layer? Density of silver is 10.5 g/cm³.

Solution:  Mass of silver to be deposited

                     = Volume × density

                     = Area ×thickness × density

Given: Area = 80 cm², thickness = 0.0005 cm and density = 10.5 g/cm³

                        Mass of silver to be deposited = 80 × 0.0005 × 10.5

                                                                = 0.42 g

                        Applying to silver E = Z × 96500

                                                Z =  108/96500 g

                        Let the current be passed for r seconds.

                        We know that

                                W = Z × I × t

                        So, 0.42 = 108/96500×3×t

                        t = (0.42 × 96500)/(108×3) = 125.09 secs

Uses of Electrolysis

Electrolysis has wide applications in industries. Some of the important applications are, as follows,

(i) Production of hydrogen by electrolysis of water.
(ii) Manufacture of heavy water.
(iii) The metals like K, Mg, Al, etc., are obtained by electrolysis of fused electrolytes.
(iv) Non-metals like hydrogen, fluorine, chlorine are obtained by electrolysis.
(v) In this method pure metal is deposited at cathode from a solution containing the metal ions etc.
(vi) Compounds like NaOH, KOH, white lead, etc. are synthesized by electrosynthesis method.
(vii) Electroplating: The process of coating an inferior metal with a superior metal by electrolysis is known as electroplating. In other word, it is the process of coating a cheaper metal with a more expensive one, such as copper or silver.

  How it works

• The negative electrode should be the object that is to be electroplated
• The positive electrode should be the metal that you want to coat the object with
• The electrolyte should be a solution of the coating metal, such as its metal nitrate or sulphate

Here are two examples:

Electroplating with silver

The object to be plated, such as a metal spoon, is connected to the negative terminal of the power supply. A piece of silver is connected to the positive terminal. The electrolyte is silver nitrate solution.

Electroplating with copper

The object to be plated, such as a metal pan, is connected to the negative terminal of the power supply. A piece of copper is connected to the positive terminal. The electrolyte is copper (II) tetraoxosulphate (VI) solution.

This arrangement can also be used to purify copper during copper manufacture. In this case, both electrodes are made from copper. The negative electrode gradually gets coated with pure copper as the positive electrode gradually disappears.

Electrolysis of brine

Brine is a solution of sodium chloride (NaCl) and water (H₂O). The process of electrolysis involves using an electric current to bring about a chemical change and make new chemicals. The electrolysis of brine is a large-scale process used to manufacture chlorine from salt. Two other useful chemicals are obtained during the process, sodium hydroxide (NaOH) and hydrogen (H₂).

It is important that the chlorine and sodium hydroxide produced in the process are separated they react when they come into contact with each other.

EXERCISES

Lets see how much you’ve learnt, attach the following answers to the comment below

1. Calculate the mass of aluminium deposited when a current of 3.0 amperes is passed through an aluminium electrolyte for 2 hours (Al = 27, 1 Faraday = 96500C) (a) 1.0g (b) 6.04g (c) 4.04g (d) 2.02g
2. In electrolytic purification processes, the impure metal to be purified is used as (a) anode (b) cathode (c) electrolyte (d) salt bridge
3. In the electrolysis of brine, the anode must be carbon because (a) carbon is a reducing agent (b) chlorine does not reduce carbon (c) carbon induces the discharge of chlorine (d) carbon is very reactive with chlorine
4. Corrosion in iron is called (a) tarnishing (b) rusting (c) electrode corrosion (d) galvanization
5. During the electrolysis of a salt of a metal M, a current of 0.5A flows for 32 minutes 10 seconds and deposits 0.325g of M. what is the charge on the metal ion? (M = 65g, 1F = 96500C) (a) 1 (b) 2 (c) 3 (d) 4

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