For my investigation, I am using copper sulphate and zinc sulphate. When the solutions are 1mol dm-3, the redox reaction between them is:
Zn(s) + CuSO4 (l) ⇌ ZnSO4 (l) + Cu(s)
The redox reaction that occurs is a displacement reaction, as it would have been carried out in one reaction vessel. The reaction would release energy as heat. However, if the two reactions were split into two half-cells, that energy can be channelled through the flow of the electrons into electricity[1].
The half-equations for the reactions are:
Zn2+(aq)|Zn(s) Zn ⇌ Zn2+ + 2e- E0 = -0.762V
Cu2+(aq)|Cu(s) Cu2+ + 2e- ⇌ Cu E0 = +0.342V
The standard electrode potential of the whole reaction is worked out by taking away the standard electrode potential of the anode, which will be the zinc, from the cathode (the copper). This means that for the ionic equation:
Zn(s) + Cu2+(aq) ⇌ Zn2+(aq) + Cu(s)
E0 = 0.342 - (-0.762) = 0.342 + 0.762 = 1.104V
Salt Bridge[3]
To complete the circuit of the cell, a salt bridge is used. The salt bridge is used to let the electrons flow from one of the solutions to another. For my investigation, I will be using filter paper soaked in saturated potassium nitrate (KNO3). Potassium nitrate is used because it conducts electricity, but it will not react with either of the reagents.
Nernst Equation
Another way of working out the electrode potential is by using the Nernst equation[4]. This method is mainly used when the electrochemical cell is not under standard conditions, such as having different concentrations. This is a simpler form of it:
E = E0 – (0.05915∕z) x log([red]∕[ox])
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E is the electrode potential.
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E0 is the standard electrode potential.
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z is the number of electrons transferred in the ionic equation.
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[red] and [ox] are the concentrations of the reduced and oxidated solutions respectively.
Plan
Equipment
- 10 x 100ml Beakers
- 2 x 50ml Measuring Cylinders
- 3 x 250ml Volumetric Flasks
- High-resistance Voltmeter
- 2 x Cables & Crocodile Clips
- Strips of Copper & Zinc
- Emery Paper
- Filter Paper
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Copper (II) Sulphate, 1 mol dm-3
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Zinc Sulphate, 1mol dm-3
- Potassium Nitrate, saturated
- Thermometer
- Scales
Risk Assessment[5]
Instructions
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Make a 1mol dm-3 solution of copper (II) sulphate and zinc sulphate.
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Pour 50cm3 of each into 100cm3 beakers. Using a folded piece of filter paper soaked in saturated potassium nitrate for the salt bridge, and a voltmeter (as shown in fig. 3), measure the electrode potential between the two solutions. Measure the temperature of the solutions at the time of the reaction.
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Repeat number 2 for all of the different concentrations (1, 0.5, 0.25, 0.125, 0.625mol dm-3)
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Keeping the copper (II) sulphate at 1mol dm-3, change the concentration of the zinc sulphate
Making the Solutions
All of the solutions that I am going to use in my investigation have to be made from the solid. Each of the compounds have different molar masses, so I have to use equations to work out how to make 1mol dm-3 of them in 250ml of water.
Copper (II) Sulphate: 1mol dm-3 for 250ml
Moles= 1 x 250 x 10-3 = 0.25mol
Molar Mass= 64 + 32 + (4 x 16) = 160g/mol
Mass= 0.25 x 160 = 40g
Zinc Sulphate: 1mol dm-3 for 250ml
Moles= 1 x 250 x 10-3 = 0.25mol
Molar Mass= 65 + 32 + (4 x 16) = 161g/mol
Mass= 0.25 x 161 = 40.25g
I was originally going to use a maximum of 2mol dm-3 for the solutions, but the maximum solubility for copper (II) sulphate is 79g for 250ml[6], which means that the concentration would not be exactly 2mol dm-3, as I would need 80g of the solid. Making the saturated potassium nitrate was easier to do because I needed at least 36g per 100ml[7].
To make a dilute concentration from the original solution, I will have to use a ratio of water to the solution.
This means that if I want to make 50ml of CuSO4 at 0.25mol dm-3, I would need to add 12.5ml of CuSO4 to 37.5ml of water.
Analysis
From the data, you can see that there is a connection between the concentration and the electrode potential. As the concentration of the copper (II) sulphate increases, the electrode potential increases. On the other hand, when the concentration of the zinc sulphate is increased, the electrode potential decreases.
You can also see that the electrode potential increases when the concentration of copper (II) sulphate decreases. The half-cell potential of copper (II) sulphate also increases when the concentration decreases. This happens because as the concentration decreases, the number of electrons in the solution is less than usual. Because of this, the solution is even more likely to receive more electrons, making it more positive. The half-cell potential of zinc sulphate also increases because it is less likely to give away electrons than usual.
In the graphs, you can see that when you compare the electrode potentials for each concentration of copper (II) sulphate, all of the lines are almost parallel to each other. This shows that if, for example, you have 1mol dm-3 of copper (II) sulphate and 0.5mol dm-3 of zinc sulphate, you can tell that the electrode potential is 1.084V. If you change the concentration of copper (II) sulphate to 0.5mol dm-3, you would have to add 0.020V to it, and then if you had 0.25mol dm-3 of copper (II) sulphate, you add 0.021V. For each one, you add around 0.020V, so you can work out the approximate electrode potential.
Evaluation
Measurement Errors
Beakers:
0.05ml∕50ml x 100 = 0.1%
0.05ml∕200ml x 100 = 0.025%
0.05ml∕250ml x 100 = 0.02%
Measuring Cylinders:
0.05ml∕50ml x 100 = 0.1%
Thermometer:
± 0.5°C
Voltmeter:
± 0.0005V
Volumetric Flask:
0.5ml∕250ml x 100 = 0.2%
Burette:
0.05ml∕25ml x 100 = 0.2%
Scales:
0.0005g∕40g x 100 = 1∕800%
0.0005g∕40.5g x 100 = 1∕810%
Bibliography
- Chemistry (Second Edition)
Ann & Patrick Fullick Heinemann, 2000
- A2 Chemistry
Andrew Hunt Hodder & Soughton, 2006
- Further Advanced Chemistry
B. Earl & LDR Wilford John Murray Publishers, 2001
- The Nernst Equation
- Hazcards
- Copper (II) sulphate – Wikipedia, the free encyclopedia
- ICSC 0184 – POTASSIUM NITRATE
Pictures:
Page 1:
Figure 1,3: Based on Further Advanced Chemistry, figure 3.10, pg. 37
Figure 2: Based on Further Advanced Chemistry, figure 3.11, pg. 38
Page 5:
