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When inert electrodes are used in the electrolysis of sodium sulfate solution, [tex]Na_2SO_4 \, \text{(aq)}[/tex], the relevant half-cell reactions are given below:

I. [tex]2H_2O \, \text{(l)} + 2e^- \rightarrow H_2 \, \text{(g)} + 2OH^- \, \text{(aq)}[/tex]
[tex]\[ E^\circ = -0.83 \, \text{V} \][/tex]

II. [tex]Na^+ \, \text{(s)} + e^- \rightarrow Na \, \text{(s)}[/tex]
[tex]\[ E^\circ = -2.71 \, \text{V} \][/tex]

III. [tex]2H_2O \, \text{(l)} \rightarrow O_2 \, \text{(g)} + 4H^+ \, \text{(aq)} + 4e^-[/tex]
[tex]\[ E^\circ = -1.23 \, \text{V} \][/tex]

IV. [tex]2SO_4^{2-} \, \text{(aq)} \rightarrow S_2O_4^{2-} \, \text{(aq)} + 2e^-[/tex]
[tex]\[ E^\circ = -2.01 \, \text{V} \][/tex]

Which of these half-cell reactions are possible cathode reactions?

A. I and II
B. I and III
C. II and IV
D. III and IV



Answer :

GinnyAnswer
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

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