Discussion Examples for Electrolysis Experiments | Electrode Reactions, Products, and Faraday’s Law

What Is Electrolysis?

Electrolysis is an operation in which direct current is passed through a solution or molten salt containing an electrolyte to cause redox reactions at electrode surfaces.
In a battery, electrical energy is extracted from a chemical reaction, whereas in electrolysis, electrical energy is supplied from an external power source to drive a chemical reaction.
Therefore, electrolysis can be understood as the reverse concept of a battery.

In electrolysis, electrons are supplied from the external power source to the cathode, so a reduction reaction occurs there.
On the other hand, electrons are removed at the anode, so an oxidation reaction occurs.
Which substance reacts changes depending on the composition of the electrolyte, electrode materials, voltage, concentration, pH, and other factors.

Main Items to Include in the Results

In the results of an electrolysis experiment, organize the type and concentration of the electrolyte, electrode materials, current, voltage, energization time, changes observed at the electrodes, generated gases, deposits, mass changes, pH changes, and other information.
When using Faraday’s law, determine the electric charge from the current and time and compare the theoretical amount of product with the measured value.

Main Items to Include in the Results

• Type of electrolyte
• Electrolyte concentration
• Electrode materials
• Changes observed at the cathode
• Changes observed at the anode
• Type of gas generated
• Presence or absence of deposited metal
• Dissolution or discoloration of the electrodes
• Current value
• Voltage
• Energization time
• Electrode masses before and after energization
• Mass or volume of products
• Theoretical amount of product
• Current efficiency
• pH change
• Causes of error and points for improvement

Reference Experimental Values and Calculation Examples for Electrolysis Experiments

Here, using electrolysis of aqueous solutions as examples, electrode reactions, products, amounts of gas generated, amounts of metal deposited, and comparison with theoretical values based on Faraday’s law are organized.

In electrolysis, electrical energy is supplied from an external power source to drive redox reactions that normally do not proceed spontaneously with ease.
A reduction reaction occurs at the cathode and an oxidation reaction occurs at the anode, and the products change depending on the electrolyte and electrode materials.

Reference Experimental Conditions

Electrolysis of Copper Sulfate Aqueous Solution

When copper sulfate aqueous solution is electrolyzed using inert electrodes, copper ions are reduced at the cathode and copper is deposited, while water is oxidized at the anode and oxygen is generated.

Measurement Results for Copper Sulfate Aqueous Solution

The following shows an example in which current was passed for 600 s and the mass of copper deposited at the cathode was measured.

Example Calculation of Electric Charge

The electric charge passed during electrolysis is determined from the product of current and time.

Electric charge Q = Current I × Time t

For sample B, the current is 0.20 A and the time is 600 s, so the electric charge is calculated as follows.

Q = 0.20 A × 600 s = 120 C

Therefore, 120 C of electric charge was passed through sample B.

Example Calculation of the Theoretical Copper Deposition Amount

Copper ions accept two electrons and are deposited as copper.

Cu2+ + 2e− → Cu

According to Faraday’s law, the theoretical deposition amount is determined using the following equation.

Theoretical deposition amount m = Q × M ÷ (n × F)

Here, Q is the electric charge, M is the molar mass of copper, 63.5 g/mol, n is the number of electrons, 2, and F is the Faraday constant, 96500 C/mol.
For sample B, Q = 120 C, so the value can be calculated as follows.

m = 120 × 63.5 ÷ (2 × 96500) = 0.0395 g

Therefore, the theoretical copper deposition amount for sample B is 0.0395 g.

Example Calculation of Current Efficiency

Current efficiency expresses as a percentage how much of the theoretical amount of product was actually obtained.

Current efficiency (%) = Measured deposition amount ÷ Theoretical deposition amount × 100

For sample B, the measured deposition amount is 0.0372 g and the theoretical deposition amount is 0.0395 g.

Current efficiency = 0.0372 ÷ 0.0395 × 100 = 94.2%

From this result, approximately 94.2% of the electric charge passed through the system was considered to have been used for copper deposition.

Electrolysis of Sodium Sulfate Aqueous Solution

In sodium sulfate aqueous solution, Na+ and SO42− are difficult to react, and mainly water is electrolyzed.
Hydrogen is generated at the cathode and oxygen at the anode.

Measurement Results for the Amounts of Gas Generated

The following shows an example in which sodium sulfate aqueous solution was electrolyzed and the amounts of hydrogen generated on the cathode side and oxygen generated on the anode side were measured.
Theoretically, hydrogen and oxygen are generated in a volume ratio of 2:1.

Example Theoretical Calculation of Hydrogen Generation

At the cathode, 1 mol of hydrogen is generated by 2 mol of electrons.

2H2O + 2e− → H2 + 2OH−

When the electric charge of the sample is 120 C, the amount of electrons is calculated as follows.

Amount of electrons = 120 C ÷ 96500 C/mol = 0.00124 mol

Because 1 mol of hydrogen is generated from 2 mol of electrons, the amount of hydrogen is calculated as follows.

Amount of hydrogen = 0.00124 ÷ 2 = 0.000622 mol

Assuming that the volume of 1 mol of gas near 25°C is approximately 24.5 L, the theoretical hydrogen volume is calculated as follows.

Hydrogen volume = 0.000622 mol × 24.5 L/mol = 0.0152 L = 15.2 mL

If the measured value is 14.2 mL, the amount of hydrogen generated is slightly smaller than the theoretical value.

Example Theoretical Calculation of Oxygen Generation

At the anode, 1 mol of oxygen is generated for every 4 mol of electrons.

2H2O → O2 + 4H+ + 4e−

When the amount of electrons is 0.00124 mol, the amount of oxygen is calculated as follows.

Amount of oxygen = 0.00124 ÷ 4 = 0.000311 mol

Using the same gas volume of 24.5 L/mol, the theoretical oxygen volume is calculated as follows.

Oxygen volume = 0.000311 mol × 24.5 L/mol = 0.00762 L = 7.62 mL

If the measured value is 6.8 mL, the smaller value compared with the theoretical value may have resulted from oxygen being more soluble in water than hydrogen or from leakage during gas collection.

Products and pH Changes Near the Electrodes

During electrolysis of water, OH− is produced near the cathode and H+ near the anode.
Therefore, when an indicator is added, color changes may be observed near the electrodes.

Differences in Products Depending on the Electrolyte

The products of electrolysis change depending on the type of electrolyte and electrode material.
Even when the same current is passed, the observed products differ depending on the ions that react more readily and the properties of the electrodes.

Example of How to Write the Results

When copper sulfate aqueous solution was electrolyzed, reddish-brown copper was deposited on the cathode surface and bubbles were generated at the anode.
Under conditions of a current of 0.20 A and an energization time of 600 s, the electric charge was 120 C.
The theoretical copper deposition amount determined from Faraday’s law was 0.0395 g, while the measured deposition amount was 0.0372 g.
Therefore, the current efficiency was calculated to be 94.2%.

In addition, when sodium sulfate aqueous solution was electrolyzed, hydrogen was generated at the cathode and oxygen at the anode.
When a current of 0.20 A was passed for 600 s, the measured hydrogen volume was 14.2 mL and the oxygen volume was 6.8 mL, giving a volume ratio of approximately 2.09:1.
This value was close to the theoretical hydrogen-to-oxygen volume ratio of 2:1 expected from electrolysis of water.

Points for Connecting the Results to the Discussion

In a discussion of electrolysis experiments, it is important to clearly identify at which electrode oxidation and reduction occurred and to relate the products to the theoretical values obtained using Faraday’s law.

• Can it be explained that reduction occurs at the cathode and oxidation at the anode?
• Can the reason why the products change depending on the electrolyte be explained?
• Has the amount of copper deposited been calculated using Faraday’s law?
• Are the amounts of hydrogen and oxygen generated close to the theoretical volume ratio of 2:1?
• Can the difference between the theoretical and measured values be explained by gas dissolution, collection leakage, side reactions, or loss of deposits?
• Can the pH changes near the electrodes be related to the production of H+ and OH−?
• Could the electrode material have participated in the reaction?

Example Discussion

In this experiment, copper sulfate aqueous solution and sodium sulfate aqueous solution were electrolyzed, and the electrode reactions and products were compared.
In the copper sulfate aqueous solution, reddish-brown copper was deposited at the cathode.
This was because Cu2+ accepted electrons and was reduced to Cu.
On the other hand, water was considered to have been oxidized at the anode, generating oxygen.

Under conditions of a current of 0.20 A and an energization time of 600 s, the electric charge was 120 C, and the theoretical copper deposition amount calculated using Faraday’s law was 0.0395 g.
The measured deposition amount was 0.0372 g, giving a current efficiency of 94.2%.
Possible reasons why the measured value was smaller than the theoretical value include side reactions other than copper deposition, loss of deposited copper, and losses during washing or drying.

In the sodium sulfate aqueous solution, hydrogen was generated at the cathode and oxygen at the anode.
The measured gas-volume ratio was H2:O2 = 2.09:1, which was close to the theoretical ratio of 2:1 for electrolysis of water.
This is because 1 mol of hydrogen is generated by 2 mol of electrons at the cathode, while 1 mol of oxygen is generated by 4 mol of electrons at the anode.

However, the measured amount of oxygen tended to be slightly smaller than the theoretical value.
Possible causes include partial dissolution of oxygen in water, leakage during gas collection, and bubbles remaining on the electrode surface.
Therefore, when discussing electrolysis results, not only the electrode reactions but also the measurement procedure and properties of the products must be considered.

Summary

In electrolysis experiments, reduction reactions occur at the cathode and oxidation reactions at the anode, and the products change depending on the electrolyte and electrode materials.

In this reference example, copper was deposited at the cathode in copper sulfate aqueous solution, while hydrogen and oxygen were generated in sodium sulfate aqueous solution.
In a report, it is useful to relate the electrode reactions, products, Faraday’s law, and the difference between theoretical and measured values.

Reactions Occurring at the Cathode

The cathode is the electrode connected to the negative terminal of the external power source.
Because electrons are supplied, a reduction reaction occurs at the cathode.
In aqueous solutions, metal ions may accept electrons and be deposited as metals, or water or hydrogen ions may be reduced and hydrogen may be generated.

For example, in an aqueous solution containing Cu2+, Cu2+ may accept electrons and be deposited as copper, Cu.
On the other hand, when ions such as Na+ that are difficult to reduce in aqueous solution are present, water may be reduced and H2 may be generated.
Which reaction occurs depends on how easily the ions are reduced and on the solution conditions.

Cu2+ + 2e- → Cu

2H2O + 2e- → H2 + 2OH-

Reactions Occurring at the Anode

The anode is the electrode connected to the positive terminal of the external power source.
Because electrons are removed, an oxidation reaction occurs at the anode.
In aqueous solutions, anions may be oxidized and generate gases, water may be oxidized and generate oxygen, or the electrode material itself may be oxidized and dissolve.

For example, in an aqueous solution containing chloride ions, Cl- may be oxidized to generate Cl2 depending on the conditions.
In an aqueous solution containing ions such as sulfate ions, SO42-, which are difficult to oxidize, water may be oxidized and O2 may be generated.
When a copper electrode is used as the anode, the copper may dissolve as Cu2+.

2Cl- → Cl2 + 2e-

2H2O → O2 + 4H+ + 4e-

Cu → Cu2+ + 2e-

How to Write Electrode-Reaction Equations

In a discussion of electrolysis, it is important to write the reactions at the cathode and anode separately.
Because the cathode involves a reduction reaction, electrons, e-, appear on the left side of the reaction equation.
Because the anode involves an oxidation reaction, electrons, e-, appear on the right side.
The position of the electrons can therefore be used to determine whether the reaction is oxidation or reduction.

When writing electrode-reaction equations, confirm that both the charge and the number of atoms are balanced.
In aqueous solutions, H+, OH-, and H2O are often involved, so the form of the equation may differ depending on whether the conditions are acidic or basic.
It is important to relate the products observed in the experiment to the reaction equations.

Points to Note in Electrolysis of Aqueous Solutions

In electrolysis of aqueous solutions, not only electrolyte ions but also water, H2O, may participate in the reaction.
Therefore, the presence of an ion in the solution does not necessarily mean that the ion will react.
At the cathode, metal ions and water may compete for reduction, while at the anode, anions and water may compete for oxidation.

For example, at the cathode during electrolysis of NaCl aqueous solution, water rather than Na+ is reduced, producing hydrogen.
This is because Na+ is difficult to deposit as metallic sodium in an aqueous solution.
In electrolysis of aqueous solutions, the products are determined by considering ionization tendency, standard electrode potentials, overvoltage, concentration, and other factors.

Discussion of Electrolysis of Water

In electrolysis of water, hydrogen, H2, is generated at the cathode and oxygen, O2, at the anode.
Because pure water conducts electricity poorly, electrolytes such as dilute sulfuric acid or sodium hydroxide may be added in experiments to increase conductivity.
The electrolyte has the role of allowing current to flow more readily and ideally is not consumed in the decomposition of water itself.

From the reaction equation, hydrogen and oxygen are generated in a molar ratio of 2:1.
If the measured gas-volume ratio differs from the theoretical value, possible causes include dissolution of the gases in water, leakage, side reactions at the electrodes, and errors in gas collection.

Cathode: 2H2O + 2e- → H2 + 2OH-

Anode: 2H2O → O2 + 4H+ + 4e-

Overall: 2H2O → 2H2 + O2

Electrolysis of Sodium Chloride Aqueous Solution

When sodium chloride aqueous solution is electrolyzed, water is reduced at the cathode to generate H2 and OH-.
At the anode, Cl- may be oxidized depending on the conditions and Cl2 may be generated.
As a result, Na+ and OH- remain in the solution, producing a state similar to sodium hydroxide aqueous solution.

However, in dilute sodium chloride aqueous solution, oxidation of water may make O2 easier to generate.
The products are affected by concentration, electrode material, overvoltage, and other factors.
Because chlorine is harmful when generated, safety precautions are extremely important.

Cathode: 2H2O + 2e- → H2 + 2OH-

Anode: 2Cl- → Cl2 + 2e-

Electrolysis of Copper Sulfate Aqueous Solution

When copper sulfate, CuSO4, aqueous solution is electrolyzed using inert electrodes, Cu2+ is reduced at the cathode and metallic copper is deposited.
At the anode, because SO42- is difficult to oxidize, water is oxidized and O2 is generated.
Therefore, reddish-brown copper may adhere to the cathode surface.

On the other hand, when a copper electrode is used as the anode, the copper at the anode is oxidized and dissolves into the solution as Cu2+.
In this case, copper is deposited at the cathode and dissolves at the anode, so the process can be understood in a manner similar to electrolytic refining or plating.
The important point is that the products change depending on the electrode material.

Cathode: Cu2+ + 2e- → Cu

Inert anode: 2H2O → O2 + 4H+ + 4e-

Copper anode: Cu → Cu2+ + 2e-

Effect of Electrode Materials

In electrolysis, the electrode material may or may not participate in the reaction.
Inert electrodes such as platinum and carbon generally do not readily participate in the reaction, so ions in the electrolyte or water mainly react.
On the other hand, metal electrodes such as copper and silver may be oxidized at the anode and dissolve into the solution.

Therefore, even with the same electrolyte, changing the electrode material changes the products and electrode mass changes.
In a report, it is necessary to clearly state not only the electrolyte but also which electrodes were used and to consider whether the electrodes participated in the reaction.

How to Confirm Products

Products generated by electrolysis can be estimated not only from observation but also by confirmation tests.
Hydrogen may burn with a small popping sound when ignited, while oxygen has the property of relighting a glowing splint.
Chlorine has a characteristic irritating odor and may turn moist potassium iodide-starch paper blue-violet.
However, gas-confirmation tests must be performed with sufficient attention to safety.

In metal deposition, the color and mass change of the electrode surface provide clues.
When copper is deposited, a reddish-brown solid may adhere to the cathode.
Confirming the products is important for determining whether the electrode-reaction equations are appropriate.

What Is Faraday’s Law?

Faraday’s law states that the amount of substance reacting during electrolysis is proportional to the electric charge passed through the system.
The electric charge Q is expressed as the product of current I and time t.
The electric charge carried by 1 mol of electrons is represented by the Faraday constant F and is approximately 96500 C/mol.

From an electrode-reaction equation, the number of moles of electrons required to obtain 1 mol of product can be determined.
For example, in Cu2+ + 2e- → Cu, 2 mol of electrons are required to deposit 1 mol of copper.
Using this relationship, the theoretical deposition amount or theoretical amount of gas generated can be calculated from the current and time.

Q = I × t

Amount of electrons = Q ÷ F

F ≒ 96500 C/mol

Discussion of Calculating Deposition Amount

When calculating the amount of metal deposited, first determine the electric charge from the current and time.
Next, use the Faraday constant to determine the amount of electrons, and convert this into the amount of metal using the number of electrons in the electrode-reaction equation.
Finally, convert this to mass using the molar mass of the metal.

For example, in copper deposition, 2 mol of electrons are required for 1 mol of Cu2+ to become Cu.
Therefore, dividing the amount of electrons passed through the system by 2 gives the theoretical amount of copper deposited.
By comparing the measured value with the theoretical value, current efficiency and side reactions can be discussed.

Cu2+ + 2e- → Cu

Amount of Cu deposited = Amount of electrons ÷ 2

Discussion of Calculating Gas Generation

When gas is generated during electrolysis, the theoretical amount can also be calculated using Faraday’s law.
Two moles of electrons are required to generate 1 mol of H2.
Four moles of electrons are required to generate 1 mol of O2.
This difference in the required number of electrons explains why H2 and O2 are generated in a ratio of 2:1 during electrolysis of water.

When comparing gas volumes, the effects of temperature, pressure, water vapor pressure, dissolution in water, and collection method must be considered.
Possible causes of differences between theoretical and measured values include gas leakage, dissolution, fluctuations in current, and errors in reading gas volume.

2H+ + 2e- → H2

2H2O → O2 + 4H+ + 4e-

Discussion of Current Efficiency

Current efficiency is a value representing the proportion of the electric charge passed through the system that was actually used for the intended reaction.
It is determined by comparing the measured amount of product with the theoretical amount.
The closer the current efficiency is to 100%, the more effectively the electric charge is considered to have been used for the intended reaction.

When current efficiency is low, possible causes include side reactions, loss of products from the electrode, gas leakage, unstable current, and poor electrode-surface condition.
In electrolysis experiments, the difference between theoretical and measured values should not simply be treated as a failure but should be discussed in terms of current efficiency and side reactions.

Current efficiency = Measured amount of product ÷ Theoretical amount of product × 100

Discussion of pH Changes

During electrolysis, H+ and OH- may be generated or consumed near the electrodes, causing the pH to change.
For example, when water is reduced at the cathode, OH- is produced and the region near the cathode becomes basic.
When water is oxidized at the anode, H+ is produced and the region near the anode may become acidic.

When an indicator is used, pH changes near the electrodes can be observed as changes in color.
However, if the entire solution is mixed well, local pH changes become more difficult to observe.
pH changes are easier to understand when related to the electrode-reaction equations.

Effects of Current and Voltage

The amount of product generated by electrolysis is proportional to the electric charge passed through the system.
Therefore, the larger the current, the greater the amount of reaction that proceeds in the same period.
However, if the voltage is too high, side reactions may become more likely, and heating or vigorous bubble generation may occur.

If the current is not constant, an error occurs in the calculation of the theoretical amount of product.
If the current changed during the experiment, the average current should be used or the changes in current should be recorded and discussed.
When bubbles adhere to the electrode surface, the effective electrode area changes and the current may become unstable.

Effect of Electrode-Surface Condition

Because electrolysis is a reaction that occurs at electrode surfaces, the condition of the electrode surfaces affects the results.
If the electrode is dirty, covered with an oxide film, covered with bubbles, or nonuniformly coated with deposits, the way current flows and the reaction rate may change.
In metal deposition, the surface condition also changes the adhesion and shape of the deposits.

When bubbles adhere to the electrode surface, the contact area with the electrolyte decreases and the reaction becomes more difficult to proceed.
In addition, if deposited metal peels off, the measured value becomes smaller in mass measurements.
It is important to clean the electrodes before the experiment and polish them when necessary.

Causes of Error in Electrolysis Experiments

Causes of error in electrolysis experiments include fluctuations in current, errors in measuring energization time, dirt on electrode surfaces, bubble adhesion, loss of products, gas leakage, dissolution of gases in water, side reactions, dissolution of electrode materials, temperature changes, changes in electrolyte concentration, and mass-measurement errors.
When Faraday’s law is used, errors in current and time directly affect the theoretical value.

If the measured amount of deposited metal is small, possible causes include loss of deposits, insufficient drying, losses during electrode washing, and side reactions.
If the measured amount of gas generated is smaller than the theoretical value, possible causes include gas leakage, dissolution in water, and errors in reading the collection vessel.
Organizing the difference between experimental and theoretical values separately for each type of product makes the discussion easier.

When the Results Can Be Considered Good

An electrolysis experiment can be considered to have produced good results when the products observed at each electrode agree with the electrode-reaction equations and the measured amount of product is close to the theoretical value determined from Faraday’s law.
In addition, if the current was stable, the energization time was accurate, and the electrode-surface condition was good, the reliability of the results is higher.

For example, if copper is deposited at the cathode in copper sulfate aqueous solution and its mass is close to the theoretical value, reduction of Cu2+ can be considered to have proceeded as the main reaction.
Likewise, if hydrogen and oxygen are generated in approximately a 2:1 ratio during electrolysis of water, the result is consistent with the reaction equations.

Example Discussion When the Experiment Did Not Go Well

When an electrolysis experiment does not go well, possible causes are considered from results such as products differing from those expected, gas volumes not agreeing with theoretical values, small amounts of metal deposition, unstable current, electrode discoloration, or pH changes not being visible.
Organizing the causes according to the electrolyte, electrode materials, current, time, confirmation of products, and measurement procedures makes the discussion easier.

How to Write Points for Improvement

In a discussion of an electrolysis experiment, writing not only the causes of error but also points for improvement makes the report easier to organize.
Points for improvement can be considered separately for the electrodes, current and time, product collection, mass measurement, and reaction conditions.

Improvements to the Electrodes and Electrolyte

• Wash the electrodes before use
• Polish the electrode surfaces when necessary
• Keep the distance between the electrodes constant
• Keep the immersed electrode areas consistent
• Prepare the electrolyte concentration accurately
• Clearly record the electrode materials

Improvements to Current and Energization Conditions

• Keep the current constant
• Record the current regularly
• Measure the energization time accurately
• Avoid excessively high voltage
• Suppress temperature increases
• Select conditions under which side reactions are less likely to occur

Improvements to Product Measurement

• Handle deposits carefully so that they do not peel off
• Wash and dry the electrodes before mass measurement
• Prevent leakage during gas collection
• Consider dissolution of gases in water
• Read gas-volume graduations correctly
• Perform multiple measurements and calculate the average value

Difference Between a Superficial Discussion and a Good Discussion

In a discussion of an electrolysis experiment, simply writing that “gas was generated” or “metal adhered” results in a superficial discussion.
A good discussion relates the reactions at the cathode and anode, electron transfer, products, Faraday’s law, and the difference between measured and theoretical values.

Examples of Expressions That Can Be Used in Reports

The following expressions can be used when writing the results and discussion of electrolysis experiments.
Adjust the necessary parts according to your own experimental results.

• In electrolysis, redox reactions are driven by supplying electrical energy from an external power source.
• A reduction reaction involving acceptance of electrons occurs at the cathode.
• An oxidation reaction involving loss of electrons occurs at the anode.
• In electrolysis of aqueous solutions, not only electrolyte ions but also water may react.
• When the electrode material participates in the reaction, the products differ from those obtained with inert electrodes.
• Generated gases and deposits must be discussed in relation to the electrode-reaction equations.
• According to Faraday’s law, the amount of product is proportional to the electric charge passed through the system.
• Electric charge is determined from the product of current and energization time.
• If the measured value is smaller than the theoretical value, side reactions or product loss may be considered.
• By determining the current efficiency, the proportion of electric charge used for the intended reaction can be evaluated.

Points to Check When Discussing Electrolysis Experiments

Checking the following points before writing the report makes the discussion easier to write.

• Are the cathode and anode distinguished?
• Is it stated that reduction occurs at the cathode and oxidation at the anode?
• Are the reaction equations for each electrode written?
• Are the generated products related to the reaction equations?
• Is the possibility that water reacts in aqueous solution considered?
• Is the effect of the electrode material considered?
• Is the theoretical amount of product calculated using Faraday’s law?
• Is the electric charge calculated from the current and energization time?
• Is the difference between measured and theoretical values discussed?
• Is current efficiency considered?
• Are side reactions and product losses explained as causes of error?
• Do the points for improvement correspond to the causes of error?

Summary

An electrolysis experiment is an experiment in which electrical energy is supplied from an external source to cause redox reactions at electrode surfaces.
A reduction reaction involving acceptance of electrons occurs at the cathode, while an oxidation reaction involving loss of electrons occurs at the anode.
In aqueous solutions, not only electrolyte ions but also water participates in reactions, so electrode-reaction equations must be carefully considered when determining the products.

Using Faraday’s law, the electric charge passed through the system can be determined from the current and energization time, and the theoretical amount of product can be calculated.
For metal deposition, the number of electrons and molar mass are considered, while for gas generation, the number of electrons and amount of gas are considered.
By comparing measured and theoretical values, current efficiency, side reactions, product losses, and measurement errors can be discussed.

In a report, rather than simply writing that “gas was generated” or “metal was deposited,” organize and discuss the reactions at the cathode and anode, electron transfer, confirmation of products, the effect of electrode materials, pH changes, Faraday’s law, current efficiency, causes of error, and points for improvement.
Electrolysis is an important experiment for specifically understanding the relationship between redox reactions and electric charge.