The cell notation Zn(s)|Zn^{2+}(aq)|Cu^{2+}(aq)|Cu(s) represents a galvanic cell operating under standard conditions - NSC Technical Sciences - Question 6 - 2020 - Paper 2
Question 6
The cell notation Zn(s)|Zn^{2+}(aq)|Cu^{2+}(aq)|Cu(s) represents a galvanic cell operating under standard conditions.
6.1.1 Define the term galvanic cell.
6.1.2 Dr... show full transcript
Worked Solution & Example Answer:The cell notation Zn(s)|Zn^{2+}(aq)|Cu^{2+}(aq)|Cu(s) represents a galvanic cell operating under standard conditions - NSC Technical Sciences - Question 6 - 2020 - Paper 2
Step 1
6.1.1 Define the term galvanic cell.
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Answer
A galvanic cell is an electrochemical cell that converts chemical energy into electrical energy. This process occurs through spontaneous redox reactions, where oxidation and reduction reactions occur in separate half-cells.
Step 2
6.1.2 Draw a labelled diagram to represent the Zn-Cu cell. Show the direction of electron flow in the external circuit.
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Answer
The labelled diagram of the Zn-Cu cell includes:
Anode (Zn electrode): where oxidation occurs, losing electrons.
Cathode (Cu electrode): where reduction occurs, gaining electrons.
Salt bridge: allows the movement of ions to maintain electrical neutrality.
Direction of electron flow: from the anode (Zn) to the cathode (Cu) through the external circuit.
Step 3
6.1.3 Write down TWO standard conditions under which the Zn-Cu cell operates.
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The temperature must be at 25°C (298 K).
The concentration of the electrolytes should be 1 mol/dm³.
Step 4
6.1.4 To which half-cell do the anions in the salt bridge migrate?
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The anions in the salt bridge migrate towards the anode half-cell (Zn). This is because the anode is losing electrons, creating a positive charge that attracts negatively charged ions.
Step 5
6.1.5 Explain the answer to QUESTION 6.1.4.
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Anions migrate towards the anode to neutralize the charge buildup that occurs as zinc is oxidized and goes into solution as Zn²⁺ ions, preventing potential buildup that would stop the cell from operating.
Step 6
6.2.1 Identify electrode X by means of a calculation.
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Answer
Given the cell potential (E°) is 2.00 V, we can assume that electrode X is a metal that, when oxidized, leads to a positive potential. Based on standard reduction potentials, electrode X is identified as Aluminum (Al).
Step 7
6.2.2 Write down the half-reaction taking place at the anode.
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At the anode, the half-reaction is:
ightarrow ext{Al}^{3+} (aq) + 3e^-$$
This reaction represents the oxidation of aluminum.
Step 8
6.2.3 Which electrode will experience a decrease in mass?
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The electrode X (Aluminum) will experience a decrease in mass, as it is oxidized, losing Al atoms to form Al³⁺ ions.
Step 9
6.2.4 Explain the answer to QUESTION 6.2.3.
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Electrode X loses mass because during oxidation, atoms of aluminum are converted into Al³⁺ ions. These ions enter the solution, reducing the amount of solid aluminum at the electrode, hence leading to a decrease in mass.
