Why Does Electrolysis of Sodium Sulphate Show Contradictory Indicator Changes?

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The electrolysis of sodium sulfate (Na2SO4) leads to contradictory indicator changes due to simultaneous reactions at the cathode and anode. At the cathode, sodium ions can be reduced to sodium metal or water can be reduced to hydrogen gas, while at the anode, sulfate ions can be oxidized to oxygen gas or water can produce oxygen and hydrogen ions. The 3V battery used is sufficient to drive these reactions, resulting in a basic indicator at the cathode and an acidic indicator at the anode. The overall reaction formula, Na2SO4 + 2H2O → 2NaOH + H2SO4, represents the most favorable products based on their electric potentials. Thus, the observed contradictions arise from the complexity of the electrolysis process and the multiple possible reactions occurring simultaneously.
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Ok this is about electrolysis of sodium Sulphate

I believe the formula is

Na2SO4 + 2H2O ----> 2NaOH + H2SO4

During the lab, the indicator at the cathode turned basic and the anode indicated acidic.

Explaining this using electric potentials:
Na + e = Na E = -2.71 V
H2O + e = ½ H2 + OH- E = -0.83 V
S2O8 + 2e = 2SO4 E = 2.05 V
½ O2 + 2H = H2O E = 1.23 V

water = 1.23 – (-0.83) = 2.06 V
sodiumSulphate = 2.05 – (-2.71) = 4.76 V

The battery used were 3V

I have no idea how this results in the original reaction formula above. Can someone englighten me? :smile:
 
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ummm i think i solved the problem, but thanks anyway.

Dunno how to delete the thread :cry:
 


It is possible for the reactions to have contradicting predictions due to the complexity of the electrolysis process. The overall reaction of electrolysis of sodium sulphate can be represented by the formula Na2SO4 + 2H2O ----> 2NaOH + H2SO4. However, during the process, multiple reactions are taking place at the cathode and anode.

At the cathode, sodium ions (Na+) are reduced to sodium metal (Na) with a potential of -2.71 V. This means that sodium metal is the most favorable product at the cathode. However, water (H2O) can also be reduced to form hydrogen gas (H2) with a potential of -0.83 V. This means that hydrogen gas is also a possible product at the cathode.

At the anode, sulfate ions (SO4) are oxidized to form oxygen gas (O2) with a potential of 2.05 V. This means that oxygen gas is the most favorable product at the anode. However, water (H2O) can also be oxidized to form oxygen gas (O2) and hydrogen ions (H+) with a potential of 1.23 V. This means that both oxygen gas and hydrogen ions are possible products at the anode.

In this scenario, the battery used has a voltage of 3V, which is higher than the potential for hydrogen gas at the cathode (-0.83 V) and lower than the potential for oxygen gas at the anode (1.23 V). This means that both hydrogen gas and oxygen gas are produced during the electrolysis process, resulting in the basic and acidic indicators at the cathode and anode, respectively.

Therefore, the contradicting predictions are a result of multiple reactions occurring simultaneously during electrolysis, and the overall reaction formula is a representation of the most favorable products based on their individual potentials.
 
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