To determine which half-cell reactions are possible cathode reactions during the electrolysis of sodium sulfate solution, we need to consider the standard electrode potentials (E°) for each reaction. In electrolytic cells, the reduction reaction takes place at the cathode, and the reaction with a more positive potential (or less negative) is favored to occur at the cathode.
Given the following reactions and their potentials:
1. [tex]\( 2 H_2O (l) + 2 e^- \rightarrow H_2(g) + 2 OH^- (aq) \)[/tex], [tex]\( E^\circ = -0.83 \text{ V} \)[/tex]
2. [tex]\( Na^+ (s) + e^- \rightarrow Na (s) \)[/tex], [tex]\( E^\circ = -2.71 \text{ V} \)[/tex]
3. [tex]\( 2 H_2O (l) \rightarrow O_2(g) + 4 H^+ (aq) + 4 e^- \)[/tex], [tex]\( E^\circ = -1.23 \text{ V} \)[/tex]
4. [tex]\( 2 SO_4^{2-} (aq) \rightarrow S_2O_4^{2-} (aq) + 2 e^- \)[/tex], [tex]\( E^\circ = -2.01 \text{ V} \)[/tex]
We look for the reactions with the highest (least negative) reduction potentials because those are the most likely to undergo reduction at the cathode.
Evaluating the potentials:
- Reaction I: [tex]\( E^\circ = -0.83 \text{ V} \)[/tex]
- Reaction II: [tex]\( E^\circ = -2.71 \text{ V} \)[/tex]
- Reaction III: [tex]\( E^\circ = -1.23 \text{ V} \)[/tex]
- Reaction IV: [tex]\( E^\circ = -2.01 \text{ V} \)[/tex]
The less negative potentials are:
- [tex]\( -0.83 \text{ V} \)[/tex] (Reaction I)
- [tex]\( -1.23 \text{ V} \)[/tex] (Reaction III)
While the more negative potentials are:
- [tex]\( -2.71 \text{ V} \)[/tex] (Reaction II)
- [tex]\( -2.01 \text{ V} \)[/tex] (Reaction IV)
Therefore, the possible cathode reactions are those with the most positive reduction potentials from the given set, which are Reaction I and Reaction II.
Hence, the correct option is:
I and II
