Experiment 1A (Textbook) – Chemistry Form 5 Chapter 1 (Redox Equilibrium)

Experiment 1A (Determining the Voltage of a Voltaic Cell):
Aim: To determine the voltage of a voltaic cell by using different pairs of metals.

Problem statement: How can different pairs of metals dipped into electrolytes affect the voltage of the voltaic cell?

Hypothesis: The greater the difference of the value of standard electrode potential of pairs of metals, the greater the voltage reading.

Variables:
(a) Manipulated variables : Different pairs of metals.
(b) Responding variables : The voltage reading of the cell.
(c) Fixed variables : Volume and concentration of electrolytes.

Operational definition – Chemical cell: Voltmeter shows a reading when different types of pairs of metals are dipped into electrolytes.


Materials: State a list of different metal plates and suitable salt solutions with a concentration of 1.0 mol dm⁻³.

Apparatus: Sandpaper, voltmeter and connecting wire with crocodile clips and state a suitable apparatus to construct a chemical cell.

Procedure:
1. Using the provided metal plates and salt solutions, plan an experiment to construct a simple chemical cell or a chemical cell that combines two half-cells.

2. Draw a labelled diagram of a simple chemical cell or a chemical cell that combines two half-cells to carry out this experiment.

3. Write the investigation steps clearly.

4. Record all observations.


Results:
Record all observations in the following table.


Discussion:
For each set of the experiment:
1. write the half equation for oxidation reaction, half equation for reduction reaction and the overall ionic equation.

2. write the cell notations for the voltaic cells.

3. calculate the theoretical voltage of the cells using the standard electrode potential value of the half-cells.

4. deduce the relationship between the pair of metals and the voltage of the cells.


Answer:
Sample answer: Simple chemical cell

Materials: Magnesium ribbon, iron nail, zinc plate, lead plate, copper plate and copper(II) nitrate solution.

Apparatus: Beaker

Procedure:
1. Using the sand paper, clean the magnesium ribbon, iron nail, zinc plate, lead plate and copper plate.

2. Pour 1.0 mol dm⁻³ of copper(II) nitrate solution into a beaker until half full.


3. Connect magnesium ribbon and copper plate to a voltmeter using a connecting wire.

4. Dip the magnesium ribbon and copper plate into the copper(II) nitrate solution to complete the circuit.

5. Record the voltmeter reading, the metal at the negative terminal and the metal at the positive terminal.

6. Repeat steps 3 to 5 using the iron nail, zinc plate and lead plate to replace magnesium ribbon.


Or
Sample answer: Chemical cell with two half cells.

Materials: Magnesium ribbon, iron nail, zinc plate, lead plate, copper plate, magnesium nitrate solution, iron(II) nitrate solution, zinc nitrate solution, lead(II) nitrate solution and copper(II) nitrate solution.

Apparatus: Beaker and porous pot

Procedure:
1. Using the sand paper, clean the magnesium ribbon, iron nail, zinc plate, lead plate and copper plate.

2. Pour 1.0 mol dm⁻³ of magnesium nitrate solution into a porous pot and 1.0 mol dm⁻³ of copper(II) nitrate solution into a beaker until half full.



3. Place the porous pot into the beaker.

4. Connect the magnesium ribbon and copper plate to a voltmeter using a connecting wire.

5. Dip the magnesium ribbon into the magnesium nitrate solution and the copper plate into the copper(II) nitrate solution to complete the circuit.

6. Record the voltmeter reading, the metal at the negative terminal and the metal at the positive terminal.

7. Repeat steps 2 to 6 using the iron(II) nitrate solution, zinc nitrate solution and lead(II) nitrate solution to replace magnesium nitrate solution in the porous pot, and using the iron nail, zinc plate and lead plate to replace magnesium ribbon.


Discussion:
1.
– Pair of Mg/Cu 
$$\begin{array}{ll} \text { Oxidation half equation } & : \mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \\ \text {Reduction half equation } & : \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \\ \text { Overall ionic equation } & : \mathrm{Mg}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Cu} \end{array}$$

– Pair of Fe/Cu
$$\begin{array}{ll} \text { Oxidation half equation } & : \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \\ \text {Reduction half equation } & : \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \\ \text { Overall ionic equation } & : \mathrm{Fe}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Fe}^{2+}+\mathrm{Cu} \end{array}$$

– Pair of Zn/Cu
$$\begin{array}{ll} \text { Oxidation half equation } & : \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \\ \text {Reduction half equation } & : \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \\ \text { Overall ionic equation } & : \mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Cu} \end{array}$$

– Pair of Pb/Cu
$$\begin{array}{ll} \text { Oxidation half equation } & : \mathrm{Pb} \rightarrow \mathrm{~Pb}^{2+}+2 \mathrm{e}^{-} \\ \text {Reduction half equation } & : \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \\ \text { Overall ionic equation } & : \mathrm{Pb}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Pb}^{2+}+\mathrm{Cu} \end{array}$$


2. 
– Pair of Mg/Cu
$$\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}\left(\mathrm{aq}, 1 \mathrm{~mol} \mathrm{dm}^{-3}\right) \| \mathrm{Cu}^{2+}\left(\mathrm{aq}, 1 \mathrm{~mol} \mathrm{dm}^{-3}\right)\right| \mathrm{Cu}(\mathrm{s})$$

– Pair of Fe/Cu
$$\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}\left(\mathrm{aq}, 1 \mathrm{~mol} \mathrm{dm}^{-3}\right) \| \mathrm{Cu}^{2+}\left(\mathrm{aq}, 1 \mathrm{~mol} \mathrm{dm}^{-3}\right)\right| \mathrm{Cu}(\mathrm{s})$$

– Pair of Zn/Cu
$$\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}\left(\mathrm{aq}, 1 \mathrm{~mol} \mathrm{dm}^{-3}\right) \| \mathrm{Cu}^{2+}\left(\mathrm{aq}, 1 \mathrm{~mol} \mathrm{dm}^{-3}\right)\right| \mathrm{Cu}(\mathrm{s})$$

– Pair of Pb/Cu
$$\mathrm{Pb}(\mathrm{s})\left|\mathrm{Pb}^{2+}\left(\mathrm{aq}, 1 \mathrm{~mol} \mathrm{dm}^{-3}\right) \| \mathrm{Cu}^{2+}\left(\mathrm{aq}, 1 \mathrm{~mol} \mathrm{dm}{ }^{-3}\right)\right| \mathrm{Cu}(\mathrm{s})$$


3. 
– Pair of Mg/Cu
$$\mathrm{E}_{\text {cell }}^0=(+0.34)-(-2.38)=+2.72 \mathrm{~V}$$

– Pair of Fe/Cu
$$\mathrm{E}_{\text {cell }}^0=(+0.34)-(-0.44)=+0.78 \mathrm{~V}$$

– Pair of Zn/Cu
$$\mathrm{E}_{\text {cell }}^0=(+0.34)-(-0.76)=+1.10 \mathrm{~V}$$

– Pair of Pb/Cu
$$\mathrm{E}_{\mathrm{cell}}^0=(+0.34)-(-0.13)=+0.47 \mathrm{~V}$$

4. The greater the difference of E⁰ value for the pair of metals, the greater the voltage of the cell.