Answer :
Let's break down the given problem into manageable parts and solve each question step by step.
### Part 1: Potassium and Iodine Reaction
Given the reaction:
[tex]\[ 2 K(s) + I_2(g) \rightarrow 2 KI(s) \][/tex]
#### 56. What element undergoes oxidation?
Oxidation is the process of losing electrons. In the reaction:
[tex]\[ 2 K(s) \rightarrow 2 K^+ + 2 e^- \][/tex]
Potassium (K) is oxidized because it loses electrons to form [tex]\( K^+ \)[/tex].
Answer: Potassium (K) undergoes oxidation.
#### 57. What element undergoes reduction?
Reduction is the process of gaining electrons. In the reaction:
[tex]\[ I_2(g) + 2 e^- \rightarrow 2 I^- \][/tex]
Iodine ([tex]\( I_2 \)[/tex]) is reduced because it gains electrons to form [tex]\( I^- \)[/tex].
Answer: Iodine ([tex]\( I_2 \)[/tex]) undergoes reduction.
#### 58. What element is the oxidizing agent?
The oxidizing agent is the species that causes oxidation by being reduced itself. Since iodine ([tex]\( I_2 \)[/tex]) gains electrons and is reduced, it is the oxidizing agent.
Answer: Iodine ([tex]\( I_2 \)[/tex]) is the oxidizing agent.
### Part 2: Voltaic Cell with Cu / [tex]\( Cu^{2+} \)[/tex] and Hg / [tex]\( Hg_2^{2+} \)[/tex]
#### Write the net reaction for this voltaic cell and calculate [tex]\( E^\circ \)[/tex] for the cell.
Given half-reactions:
[tex]\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E^\circ = +0.34 \, \text{V} \][/tex]
[tex]\[ \text{Hg}_2^{2+} + 2e^- \rightarrow 2 \text{Hg} \quad E^\circ = +0.79 \, \text{V} \][/tex]
To determine which half-reaction is the anode and which is the cathode, remember that the cathode is where reduction occurs and the anode is where oxidation occurs.
- The more positive potential corresponds to the reduction at the cathode.
- The less positive potential corresponds to the oxidation at the anode.
In this case:
- [tex]\( \text{Hg}_2^{2+} + 2e^- \rightarrow 2 \text{Hg} \)[/tex] will occur at the cathode (higher potential, [tex]\( E^\circ = +0.79 \, \text{V} \)[/tex])
- [tex]\( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)[/tex] will occur at the anode (lower potential, [tex]\( E^\circ = +0.34 \, \text{V} \)[/tex])
To find the standard cell potential [tex]\( E^\circ_{\text{cell}} \)[/tex], we use the equation:
[tex]\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \][/tex]
So we have:
[tex]\[ E^\circ_{\text{cell}} = +0.79 \, \text{V} - (+0.34 \, \text{V}) = +0.45 \, \text{V} \][/tex]
Net reaction:
[tex]\[ 2 \text{Cu} + \text{Hg}_2^{2+} \rightarrow 2 \text{Cu}^{2+} + 2 \text{Hg} \][/tex]
Answer:
The net reaction for the voltaic cell is:
[tex]\[ 2 \text{Cu} + \text{Hg}_2^{2+} \rightarrow 2 \text{Cu}^{2+} + 2 \text{Hg} \][/tex]
The standard cell potential [tex]\( E^\circ_{\text{cell}} \)[/tex] is:
[tex]\[ +0.45 \, \text{V} \][/tex]
This solution addresses each part of the problem in a structured manner, ensuring clarity in explaining the oxidation-reduction processes, identifying the oxidizing agent, and calculating the cell potential.
### Part 1: Potassium and Iodine Reaction
Given the reaction:
[tex]\[ 2 K(s) + I_2(g) \rightarrow 2 KI(s) \][/tex]
#### 56. What element undergoes oxidation?
Oxidation is the process of losing electrons. In the reaction:
[tex]\[ 2 K(s) \rightarrow 2 K^+ + 2 e^- \][/tex]
Potassium (K) is oxidized because it loses electrons to form [tex]\( K^+ \)[/tex].
Answer: Potassium (K) undergoes oxidation.
#### 57. What element undergoes reduction?
Reduction is the process of gaining electrons. In the reaction:
[tex]\[ I_2(g) + 2 e^- \rightarrow 2 I^- \][/tex]
Iodine ([tex]\( I_2 \)[/tex]) is reduced because it gains electrons to form [tex]\( I^- \)[/tex].
Answer: Iodine ([tex]\( I_2 \)[/tex]) undergoes reduction.
#### 58. What element is the oxidizing agent?
The oxidizing agent is the species that causes oxidation by being reduced itself. Since iodine ([tex]\( I_2 \)[/tex]) gains electrons and is reduced, it is the oxidizing agent.
Answer: Iodine ([tex]\( I_2 \)[/tex]) is the oxidizing agent.
### Part 2: Voltaic Cell with Cu / [tex]\( Cu^{2+} \)[/tex] and Hg / [tex]\( Hg_2^{2+} \)[/tex]
#### Write the net reaction for this voltaic cell and calculate [tex]\( E^\circ \)[/tex] for the cell.
Given half-reactions:
[tex]\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E^\circ = +0.34 \, \text{V} \][/tex]
[tex]\[ \text{Hg}_2^{2+} + 2e^- \rightarrow 2 \text{Hg} \quad E^\circ = +0.79 \, \text{V} \][/tex]
To determine which half-reaction is the anode and which is the cathode, remember that the cathode is where reduction occurs and the anode is where oxidation occurs.
- The more positive potential corresponds to the reduction at the cathode.
- The less positive potential corresponds to the oxidation at the anode.
In this case:
- [tex]\( \text{Hg}_2^{2+} + 2e^- \rightarrow 2 \text{Hg} \)[/tex] will occur at the cathode (higher potential, [tex]\( E^\circ = +0.79 \, \text{V} \)[/tex])
- [tex]\( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)[/tex] will occur at the anode (lower potential, [tex]\( E^\circ = +0.34 \, \text{V} \)[/tex])
To find the standard cell potential [tex]\( E^\circ_{\text{cell}} \)[/tex], we use the equation:
[tex]\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \][/tex]
So we have:
[tex]\[ E^\circ_{\text{cell}} = +0.79 \, \text{V} - (+0.34 \, \text{V}) = +0.45 \, \text{V} \][/tex]
Net reaction:
[tex]\[ 2 \text{Cu} + \text{Hg}_2^{2+} \rightarrow 2 \text{Cu}^{2+} + 2 \text{Hg} \][/tex]
Answer:
The net reaction for the voltaic cell is:
[tex]\[ 2 \text{Cu} + \text{Hg}_2^{2+} \rightarrow 2 \text{Cu}^{2+} + 2 \text{Hg} \][/tex]
The standard cell potential [tex]\( E^\circ_{\text{cell}} \)[/tex] is:
[tex]\[ +0.45 \, \text{V} \][/tex]
This solution addresses each part of the problem in a structured manner, ensuring clarity in explaining the oxidation-reduction processes, identifying the oxidizing agent, and calculating the cell potential.
