In the zinc-copper cell, the zinc plate dissolves in the acid and bubbles of hydrogen gas form at the copper electrode. The overall cell reaction for this simple cell is:
Zn(s) + H2SO4 ZnSO4 (aq) + H2(g)
Oxidation and reduction occur together in order to cause the flow of electrons in a simple cell. Thus, electrical energy is produced by redox reactions in the simple cell.
How do you determine if a particular electrode is positive r negative in a simple cell?
More reactive metals tend to give up electrons and form ions more readily than less reactive metals. Thus, in a simple cell, the flow of electrons is always from the more reactive metals to the less reactive metal. The more reactive metal becomes the anode and the cell reactive becomes the cathode.
At the zinc electrode:
Oxidation occurs. Zinc atoms give up electrons and go into solution as zinc ions. Zn(s) Zn2+ + 2e- (oxidation)
The electrons leave the zinc plate, pass through the voltmeter and enter the copper plate.
At the copper electrode: The electrons are taken up by the positively charged hydrogen ions to form hydrogen gas. This is a reduction reaction.
2H+(aq) + 2e- H2(g) (reduction)
The electrode from which electrons flow out of is the negative electrode. The electrode into which the electrons flow is a positive electrode. In the simple cell, zinc acts as the negative and electrode and copper as the positive electrode. The electrolyte is a dilute hydrochloric acid. The electrons which flow in the external circuit constitute the electric current.
What happens in a zinc-copper cell when the electrolyte is copper (II) sulphate?
The figure below shows as zinc-copper cell that uses a 1.0 mol/dm3 copper (II) sulfate solution which acts as the electrolyte.
The ionic equation for the overall reaction is obtained by adding the half ionic equations for the reactions at the electrodes.
Overall ionic equation: Zn(s) + Cu2+ (aq) Zn2+(aq) + Cu(s)
Salt bridge simple cell:
A kind of simple cell that uses two different metals and two different electrolyte solutions, connected by a salt bridge is shown in the figure below.
Here the electrolytes are salt solutions of the electrode metals. Similarly, zinc becomes the anode and copper, the cathode.
At the zinc electrode: Oxidation occurs. Zinc atoms give up electrons and go into solution as zinc ions with the electrons flowing through the external set up to the copper cathode. Zn(s) Zn2+ + 2e- (oxidation)
At the cathode, each copper(II) ions from the copper(II) sulfate solution gains 2 electrons and is deposited as the copper on cathode.
The overall reaction:
Zn(s) + Cu2+ (aq) Zn2+(aq) + Cu(s) At the zinc electrode:
Zinc dissolves to form zinc ions. Zn(s) Zn2+ + 2e- (oxidation)
At the copper electrode:
Copper (II) ions from the solution receive electrons to form copper. Cu2+ + 2e- --> Cu(s)
Simple cells and reactivity series:
The reactions of metals with water and acids are used to place metals in order of their reactivity. Simple cells can also be used to determine the relative positions of metals in the reactivity series. This is because the amount of electrical energy produced in a simple cell is determined by how far apart the metals used (as electrodes) are in the reactivity series.
By using different metals in the simple cell, different voltages are produced.
METAL ELECTRODES VOLTAGE (V)
Magnesium/ Copper 2.7 Aluminium/ Copper 2.0 Zinc/ Copper 1.1 Iron/Copper 0.8 Lead/Copper 0.5 Copper/ Copper 0.0
The voltage of magnesium-copper cell is 2.7 V. If the magnesium is replaced by zinc, a less reactive metal, the voltage decreases to 1.1 V. This shows that the further apart the two metals are in the reactivity series, the greater the voltage produced. No current will flow if both electrodes are made up of the same metal.
Faraday’s law of electrolysis:
The mass of the substance liberated during electrolysis is directly proportional to the quantity of electricity produced.
The quantity of electricity required to deposit one mole of any substance is a whole number of mole of electrons.
Electrolysis Calculations:
The amount of product of an electrolysis reaction can be calculated and depends on 3 factors. The amount of charge passed which is expressed in a number of faraday.
1 faraday = the charge of 1 mole of electrons = 96 500 coulombs (C)
The amount of charge which passes in a circuit depends on the current (amperes or A) of the circuit and the amount of time (in seconds) the current is switch on.
The number of faraday can be calculated by :
o multiplying the amperes by seconds which gives coulombs (C) o dividing the number of coulombs by 96 500 C
charge of the ions: the greater the charge of the ion, the greater the number of electrons needed or the greater the number of faraday
Na+ + e- Na this means that to produce 1 mole of sodium atoms (i.e. 23 grams) one mole of electrons or 1 faraday is needed.
Ca2+ + 2e- Ca this means that to obtain 1 mole of calcium atoms two moles of electrons or 2 faraday are needed
2Cl- (aq) Cl2 (g) + 2e- this means that to obtain 1 mole of chlorine molecules two moles of electrons or 2 faraday are needed
4OH- (aq) O2 (g) + 2H2O (l) + 4e- this means that to obtain 1 mole of oxygen molecules 4 moles of electrons or 2 faraday are needed
Worked example
A solution of copper (II) sulphate is electrolysed. How much copper will be deposited by a current of 2 A flowing for 20 minutes?
Step 1: calculate the amount of coulombs (=charge)
charge = 2 A x 1200 seconds = 2400 C
Step 2: calculate the number of faraday:
faraday = 2400/96 500 = 0.025 faraday or number of moles of electrons
Step 3: use ratio of redox equation Cu2+ + 2e- Cu
2 moles of electrons give 1 mole of copper
0.025 moles of copper give 0.0125 moles of copper
How to interpret electrode equations? Moles of electrons:
Magnesium is manufactured by electrolysing molten magnesium chloride. Magnesium is produced at the cathode (the negative electrode) and chlorine at the anode (the positive electrode).The electrode
equations:
Mg2++ 2e- Mg(l) 2Cl-(l) Cl2 + 2e- In terms of moles:
1 mole of Mg2+ ions gains 2 moles of electrons and produces 1 mole of magnesium, Mg
2 moles of Cl- ions form 1 moles of chlorine, Cl2 and releases 2 moles of electrons. When we are doing calculations, we just read e- as ‘1 mole of electrons’
Quantities of electricity
The coulomb is the measure of quantity of electricity. 1 coulomb is the quantity of electricity which passes f 1 ampere (amp) flows for 1 second.
Number of coulombs = current in amps x time in seconds
So, if 2 amps flow for 20 minutes, we can calculate the quantity of electricity (1st convert the time into seconds) as:
Quantity of electricity = 2 x 20 x 60 = 2400 coulombs The faraday constant:
A flow of electricity is a flow of electrons. 1 faraday is the quantity of electricity which represents 1 mole of electrons passing a particular point in the circuit – in other approximately 6 x 1023 electrons.
1 faraday = 96, 500 coulombs. Interpreting electrode equations:
In electrolysis calculations, we are usually interested in the quantity of electricity and the mass or volume of the product. For example:
Na+ (l) + e- Na (l)
1 mole of sodium, Na, is produced by the flow of 1 mole of electrons (= 1 faraday) Cu2+ + 2e- --> Cu(s)
It takes twice as much as electricity to produce a mole copper as it does a mole of sodium. That’s because the Cu2+ ion carries twice the charge, and needs twice as many electrons to neutralise it. 2Cl- (l) Cl2 (g) + 2e-
1 mole of chlorine is produced when 2 moles of electrons (= 2 faradays) flow around the circuit. Some sample calculations:
Electrolysing copper(II) sulfate solution:
What mass of copper is deposited on the cathode during electrolysis of copper(II) sulfate solution if 0.15 amps flow for 10 minutes?
The electrode equation is: Cu2+ + 2e- --> Cu(s)
(RAM: Cu= 64, 1 faraday = 96000 coulombs)
Start by working out the number of coulombs = amps x time in seconds = 0.15 x10 x 60
= 90 coulombs Now work from the equation:
Cu2+ + 2e- --> Cu(s)
2 moles of electrons give 1 mole of copper, Cu 2 x 96000 coulombs to give 64g of copper 192,000 coulombs give 64g of copper 90 coulombs give = 0.03 g
Calculations involving gases
During electrolysis of dilute sulfuric acid using platinum electrodes, hydrogen is released at the cathode and oxygen at the anode. Calculate the volumes of hydrogen and oxygen produced (measured at room temperature and pressure) if 1.0 amp flows for 20 minutes.
The electrode equations are: 2H+ (aq) + 2e- H2 (g)
4OH- (aq) O2 (g) + 2H2O (l) + 4e-
(The molar volume of a gas = 24,000 cm3 at rtp; 1 faraday = 96000 coulombs) Number of coulombs = amps x time in seconds
= 1.0 x 20 x 60 = 1200
Calculating the volume of hydrogen:
2H+ (aq) + 2e- H2 (g)
2 moles of electrons give 1 mole of hydrogen, H2
2 x 96000 coulombs give 24,000 cm3 of hydrogen at rtp. 192, 000 coulombs give 24,000 cm3 of hydrogen at rtp. 1200 coulombs give 3
Calculating the volume of oxygen:
4OH- (aq) O2 (g) + 2H2O (l) + 4e-
A flow of 4 moles of electrons produces 1 mole of oxygen, O2 4 x 96000 coulombs produces 24000 dm3 of oxygen
384,000 coulombs produces 24,000 dm3 of oxygen 1200 coulomb produces 3
A reversed calculation:
How long will it take to deposit 0.500g of silver on the cathode during electrolysis of silver (I) nitrate solution using a current of 0.250 amps? The cathode equation is:
Ag+ (aq) + e- Ag(s)
(RAM: Ag = 108; 1 faraday = 96000 coulombs) 1 mole of electrons gives 1 mole of silver, Ag. 96000 coulombs give 108g of silver.
To produce 1g of silver we’d need = 888.89 coulombs
To produce 0.500g of silver we’d need Number of coulombs = amps x time in seconds
444.4 = 0.250 x t
T =
The time needed to deposit 0.500g of silver is 1780 seconds/ 29.67 minutes. Electrolysing more than one solution:
Suppose we have two solutions connected together in series, so that the same quantity of electricity flows through both.
At the end of the electrolysis, it was found that 2.07g of lead had been deposited on the cathode in the left-hand beaker.
a) Calculate the quantity of electricity that passes during the experiment.
b) Calculate the mass of silver that was deposited on the cathode in the right hand beaker. (RAMs: Pb= 207; Ag= 108; 1 faraday =96000 coulombs)
(a) Pb2++ 2e- Pb(s)
2 moles of electrons give 1 mole of lead, pb 2 x 96000 coulombs give 207 g of lead 192, 000 coulombs give 207 g of lead 1920 coulombs give 2.07 gram of lead.
The quantity of electricity that passed = 1920 coulombs.
(b) If 1920 coulombs passed through the beaker containing the lead(II) nitrate solution, then exactly the same amount passes through the rest of the circuit.
Ag+ (aq) + e- Ag(s)
1 mole of electrons gives 1 mole of silver, Ag 96000 coulombs give 108g of silver.
1920 coulombs give The mass of silver deposited is 2.16g.
Alternative way of solving part (b):
If we weren’t asked to find the quantity of electricity in part (a), we could do part (b) much more easily without knowing anything at all about the faraday’s constant or even about coulombs.
Equations: Pb2++ 2e- Pb(s) Ag+ (aq) + e- Ag(s)
If 2 moles of electrons flow, we will get 1 mole of lead and 2 moles of silver. However many electrons flow, we will always get twice as much as moles of silver as lead.
In this calculation, 2.07g of lead were formed, which is 0.01 mol We will therefore get 0.02 mol of silver = 0.02 x 108 = 2.16g
A similar example involving just one solution
During the electrolysis of concentrated copper(II) chloride solution, 3.2g of copper was deposited at the cathode. What volume of chlorine (measured as rtp) would be formed at the anode? (RAM: Cu = 64 ; molar volume = 24,000 cm3 at rtp)
The electrode equations are: Cu2+ + 2e- --> Cu(s)
2Cl- (l) Cl2(g) + 2e-
For every 2 moles of electrons that flow, we will get 1 mole of Copper, Cu and 1 mole chlorine, Cl2. We are bond to get the same number of moles of each.
In this case, 3.2g of copper is mol = 0.05 mol.
So we will also get 0.05 mol of chlorine which is 0.05 x 24, 000 cm3 at rtp. The volume of chlorine produced is 1200 cm3.