Potassium metal reacts with gaseous iodine (I₂) to form ionic potassium iodide according to the equation:
[tex]\[ 2 K (s) + I_2 (g) \longrightarrow 2 KI (s) \][/tex]

56. What element undergoes oxidation?

57. What element undergoes reduction?

58. What element is the oxidizing agent?

A voltaic cell has Cu/Cu²⁺ and Hg/Hg₂²⁺ half cells. [tex]\( E^{\circ} \)[/tex] is +0.34 V for the reaction [tex]\( Cu^{2+} + 2 e^{-} \longrightarrow Cu \)[/tex], and [tex]\( E^{\circ} \)[/tex] is +0.79 V for the reaction [tex]\( Hg_2^{2+} + 2 e^{-} \longrightarrow 2 Hg \)[/tex].

Write the net reaction for this voltaic cell and calculate [tex]\( E^{\circ} \)[/tex] for the cell. Use proper significant figures and units.



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.