8.1 Give a reason why platinum is used as the electrode in half-cell A - NSC Physical Sciences - Question 8 - 2019 - Paper 2
Question 8
8.1 Give a reason why platinum is used as the electrode in half-cell A.
8.2 Write down the:
8.2.1 Energy conversion that takes place in this cell.
8.2.2 Half-reac... show full transcript
Worked Solution & Example Answer:8.1 Give a reason why platinum is used as the electrode in half-cell A - NSC Physical Sciences - Question 8 - 2019 - Paper 2
Step 1
Give a reason why platinum is used as the electrode in half-cell A.
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Answer
Platinum is used as the electrode in half-cell A because it is an excellent conductor of electricity and is chemically inert. This means it does not participate in the reactions occurring at the electrode surface, thereby providing a stable interface for the electrochemical reactions.
Step 2
8.2.1 Energy conversion that takes place in this cell.
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The energy conversion that takes place in this electrochemical cell is the conversion of chemical energy into electrical energy. This occurs as redox reactions unfold, allowing charge transfer and current generation.
Step 3
8.2.2 Half-reaction that takes place at the cathode.
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The half-reaction that takes place at the cathode is:
ightarrow ext{Cr}(s)$$
Step 4
8.2.3 Cell notation for this cell.
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The cell notation for this electrochemical cell is:
extCr(s)∣extCr3+(aq)(1 mol-dm3)∣∣extCl2(g)∣extCl−(aq)(1 mol-dm3)∣extPt
Step 5
8.3 Calculate the initial emf of this cell.
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Answer
To calculate the initial emf of the cell, we can use the standard electrode potentials (E) for both half-cells. The emf can be calculated as follows:
E=Ecathode−Eanode
From the provided values:
E_{cathode} (for Cr) = -0.74 V
E_{anode} (for Cl) = +1.36 V
Thus, the calculation will yield:
E=(1.36V)−(−0.74V)=2.10V
Step 6
8.4 What will be the effect on the cell potential when a small amount of silver nitrate solution, AgNO₃(aq), is added to half-cell A?
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The addition of silver nitrate will increase the concentration of the Ag⁺ ions in half-cell A. Silver ions can react, leading to changes in the oxidation-reduction dynamics at the electrode interface. This would typically result in an increase in the cell potential.
