Question

a. In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), what reactions occur at the anode and the cathode (assuming standard conditions)? b. When water containing a small amount \((-0.01 M)\) of sodium sulfate is electrolyzed, measurement of the volume of gases generated consistently gives a result that the volume ratio of hydrogen to oxygen is not quite \(2: 1 .\) To what do you attribute this discrepancy? Predict whether the measured ratio is greater than or less than \(2: 1 .\) (Hint: Consider overvoltage.)

Short answer

During the electrolysis of an aqueous solution of Na₂SO₄, the predominant reactions that occur at the anode and cathode in standard conditions are: Anode (oxidation): Oxidation of water (formation of oxygen gas) \[ 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- \] Cathode (reduction): Reduction of water (formation of hydrogen gas) \[ 2\mathrm{H}_2\mathrm{O} + 2e^- \rightarrow 2\mathrm{OH}^- + \mathrm{H}_2 \] When water containing a small amount of sodium sulfate is electrolyzed, the measured volume ratio of hydrogen to oxygen is not quite 2:1 due to the overvoltage effect, which makes the generation of hydrogen gas slower than expected. As a result, the measured volume ratio of hydrogen to oxygen will be less than 2:1.

Step-by-step solution

To begin, we must determine the possible half-reactions that might occur at the anode (oxidation) and cathode (reduction) during the electrolysis of an aqueous solution of Na₂SO₄. Possible anode half-reactions: 1. Oxidation of sulfate ions: \[ \mathrm{SO}_4^{2-} \rightarrow \mathrm{S}_2\mathrm{O}_8^{2-} + 2e^- \] 2. Oxidation of water (formation of oxygen gas): \[ 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- \] Possible cathode half-reactions: 1. Reduction of sodium ions: \[\mathrm{Na}^+ + e^- \rightarrow \mathrm{Na}\] 2. Reduction of water (formation of hydrogen gas): \[ 2\mathrm{H}_2\mathrm{O} + 2e^- \rightarrow 2\mathrm{OH}^- + \mathrm{H}_2 \] #Step 2: Choose predominant half-reactions#

Considering the standard conditions and knowing the reactivity of the species, we can conclude the predominant reactions at the anode and cathode. Anode (oxidation): Oxidation of water (formation of oxygen gas) \[ 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- \] Cathode (reduction): Reduction of water (formation of hydrogen gas) \[ 2\mathrm{H}_2\mathrm{O} + 2e^- \rightarrow 2\mathrm{OH}^- + \mathrm{H}_2 \] #Step 3: Analyze the volume ratio discrepancy between hydrogen and oxygen generated#

We are given that the volume ratio of hydrogen and oxygen generated is consistently not quite 2:1, and we must determine whether this discrepancy results in the measured ratio being greater than or less than 2:1. When water containing a small amount of sodium sulfate is electrolyzed, the volume of hydrogen gas generated at the cathode should, theoretically, be twice the volume of oxygen gas generated at the anode, according to stoichiometry. However, in reality, the electrolysis process is affected by other factors, such as overvoltage. Overvoltage occurs when the actual voltage required to drive an electrochemical reaction exceeds the theoretical voltage (from standard electrode potentials). Since the overvoltage for the reduction of water to hydrogen gas is typically higher than the overvoltage for the oxidation of water to oxygen gas, the formation of hydrogen gas requires a higher voltage and can be slower than expected. #Step 4: Predict whether the measured ratio is greater than or less than 2:1#

Due to the overvoltage effect, the generation of hydrogen gas at the cathode can be slower than expected, leading to a lesser amount of hydrogen gas being generated as compared to the theoretical value. Therefore, the measured volume ratio of hydrogen to oxygen will be less than 2:1.

Key concepts

Electrochemical Reactions

Electrochemical reactions are at the heart of processes such as electrolysis, where electrical energy is used to drive chemical changes. This type of reaction is a cornerstone of electrochemistry and involves the transfer of electrons between a cathode and an anode, resulting in oxidation and reduction reactions.
During the electrolysis of aqueous sodium sulfate (\text( Na}_2\text(SO}_4)), we analyze possible reactions at both electrodes. At the anode, water molecules can be oxidized, releasing oxygen gas, protons, and electrons. At the cathode, water is also reduced, forming hydroxide ions and hydrogen gas. These reactions can be represented by:

• Anode (oxidation): \[ 2\text(H}_2\text(O} \rightarrow \text(O}_2 + 4\text(H}^+ + 4e^- \]
• Cathode (reduction): \[ 2\text(H}_2\text(O} + 2e^- \rightarrow 2\text(OH}^- + \text(H}_2 \]

These reactions showcase the movement of electrons and the transformation of chemical species, which is characteristic of electrochemical processes.

Overvoltage Effect

The overvoltage effect is a phenomenon that can influence the outcomes in electrolysis, impacting the efficiency and rate of electrochemical reactions. Overvoltage or overpotential refers to the extra voltage that is necessary for a reaction to occur at the electrode beyond the thermodynamically predicted voltage.
This effect is significant because it implicates that the actual electricity consumption can be higher than the one predicted by standard conditions, and it can cause certain reactions to proceed more slowly than others. During electrolysis, the overvoltage for the evolution of hydrogen at the cathode is generally greater than for the evolution of oxygen at the anode. This disparity means that generating hydrogen is a less efficient process and might be delayed.
Consequences of overvoltage are particularly relevant when considering energy efficiency for industrial processes, but also for determining the product distribution in laboratory experiments, such as the one involving the electrolysis of aqueous solutions that we are assessing.

Gas Volume Ratio

Understanding the gas volume ratio that results from electrolysis is crucial since it indicates the quantities of substances produced. According to ideal stoichiometric calculations, the reaction at the cathode produces hydrogen gas, while oxygen is formed at the anode in a 2:1 ratio, given that two water molecules are required to generate one molecule of oxygen and two molecules of hydrogen.
However, when an experiment yields a hydrogen to oxygen volume ratio that is inconsistent with this theoretical ratio, this discrepancy can often be attributed to practical considerations such as overvoltage. As mentioned earlier, the higher overvoltage for hydrogen means it is produced slower than expected, leading to a real-world ratio that deviates from the predicted 2:1.
The detailed step-by-step solution highlighted that this discrepancy might result in a measured ratio of hydrogen to oxygen that is less than 2:1. It demonstrates how the practical aspects of chemistry, like overvoltage, can affect the theoretical predictions, offering students deeper insights into the complexities of chemical reactions.