Recognize when predicted electrochemical reactions do not occur

Alberta Grade 12 Chemistry

Electrochemical Changes

Alberta Chemistry 30 — Unit B2: Electrochemical Cells (Outcome 30-B2.4k)

You have learned to predict electrochemical reactions by comparing standard reduction potentials. A reaction with a positive cell voltage is predicted to be spontaneous. An electrolytic cell is predicted to force a non-spontaneous reaction by supplying external voltage. These predictions work most of the time — but not always. In real laboratories and industries, chemists are sometimes surprised: the reaction they calculated should happen simply does not happen, or a completely different reaction occurs instead. This lesson explains the three main reasons why predicted electrochemical reactions can fail: overvoltage (overpotential), passivation of an electrode, and the presence of a competing reaction that is easier to drive. Understanding these situations is an essential skill for interpreting real electrochemical data and for answering Alberta Chemistry 30 diploma-style questions.

What you will learn

Explain why a positive predicted cell voltage does not always guarantee that a reaction will occur.
Identify the conditions — overvoltage, passivation, and competing reactions — that cause predicted electrochemical reactions to fail.
Use standard reduction potentials from the Alberta Chemistry 30 Data Booklet to predict which reaction should occur, and then evaluate whether a listed condition prevents it.
Apply these ideas to both galvanic and electrolytic cell scenarios at the Grade 12 level.

Prerequisite Review: How We Predict Electrochemical Reactions

Before exploring failures, recall how predictions are made. Every half-reaction has a standard reduction potential, E°, measured in volts (V) at SATP (298 K, 100 kPa, 1 mol/L concentrations). The Alberta Chemistry 30 Data Booklet lists these values. A more positive E° means a stronger tendency to gain electrons (be reduced).

For a galvanic (voltaic) cell, the half-reaction with the higher E° runs as a reduction at the cathode; the other runs in reverse as oxidation at the anode. The predicted cell voltage is calculated as:

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

. A positive

E°_{\text{cell}}

predicts a spontaneous reaction.

For an electrolytic cell, an external power source forces a non-spontaneous reaction. The minimum voltage needed is the magnitude of the negative

E°_{\text{cell}}

. If you supply exactly that voltage, the reaction is predicted to begin. These predictions assume ideal, standard conditions — and that is exactly where real situations can differ.

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

Standard reduction potentials (E°) are listed in the Alberta Chemistry 30 Data Booklet.
$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$ ; a positive value predicts spontaneity.
Electrolytic cells require a supplied voltage at least equal to the magnitude of the negative $E°_{\text{cell}}$ .
These predictions assume standard conditions and ideal electrode behaviour — real systems can deviate.

Overvoltage: The Extra Push a Real Reaction Needs

Imagine filling a bathtub. The water flows as predicted — but only after you turn the tap far enough past the point where it just barely opens. In electrochemistry, a similar extra push is called overvoltage (also called overpotential). Overvoltage is the additional voltage, beyond the predicted minimum, that must actually be applied to an electrolytic cell before a particular half-reaction begins at a detectable rate.

At the particle level, overvoltage arises because real electrode surfaces have energy barriers that slow down electron transfer. The standard reduction potential gives the thermodynamic prediction (the minimum energy needed), but the actual electrode process also has a kinetic barrier — think of it as friction that must be overcome. The Alberta Chemistry 30 Data Booklet and course do not require you to calculate overvoltage values; you are expected to recognize when it is relevant.

The most important example in Chemistry 30 is the electrolysis of water. The predicted minimum voltage to electrolyze water is about 1.23 V based on standard reduction potentials. In practice, no bubbles appear at 1.23 V. A voltage of roughly 1.8 V or more is typically needed before hydrogen gas and oxygen gas are produced at a meaningful rate. The gap — about 0.6 V — is the overvoltage. If you supply only 1.23 V, the predicted reaction does not occur at a useful rate.

Overvoltage is especially large for gas-producing reactions (like

\text{O}_2

and

\text{H}_2

evolution) and depends on the electrode material. Smooth platinum electrodes have lower overvoltage than many others. This means the same reaction may or may not occur depending on what the electrodes are made of, even when the applied voltage is theoretically sufficient.

V_{\text{applied}} \geq |E°_{\text{cell}}| + V_{\text{overvoltage}}

Overvoltage is the extra voltage needed beyond the predicted minimum before a reaction actually occurs at a useful rate.
It arises from kinetic barriers at real electrode surfaces, not from errors in the reduction potential tables.
Reactions that produce gases (especially $\text{O}_2$ and $\text{H}_2$ ) tend to have high overvoltages.
If the applied voltage does not exceed the sum of the predicted minimum and the overvoltage, the predicted reaction does not occur.

Passivation: When a Protective Layer Stops the Reaction

A second reason a predicted reaction fails is passivation. Passivation occurs when a solid, tightly bonded product forms directly on the surface of an electrode and physically blocks further reaction. Think of it like a coat of armour: the metal underneath may be reactive, but the outer layer seals it off from any further contact with the solution.

At the particle level, passivation works like this: the first few atoms at the electrode surface react as predicted, but the product — often a metal oxide or an insoluble salt — does not dissolve or fall away. Instead, it sticks firmly to the surface and covers the reactive metal beneath. Ions in solution can no longer reach the metal, so electron transfer effectively stops even though the reaction is still predicted to be spontaneous.

A classic example is aluminium metal in contact with oxygen. The data booklet predicts that aluminium should be oxidized readily (its E° for the

\text{Al}^{3+}/\text{Al}

half-reaction is very negative, indicating aluminium is easily oxidized). Yet a sheet of aluminium foil does not visibly corrode in air. The reason is that the surface instantly forms a thin, hard layer of aluminium oxide,

\text{Al}_2\text{O}_3\text{(s)}

, which is impermeable. The predicted corrosion reaction is blocked by its own product.

Another relevant case in electrolysis is lead. When a lead electrode is used in a sulfate-containing solution, the predicted oxidation of lead can be stopped by the formation of insoluble lead(II) sulfate,

\text{PbSO}_4\text{(s)}

, on the electrode surface. Passivation is not a failure of the prediction's logic — it is a real physical barrier that the prediction does not account for.

4\,\text{Al}\text{(s)} + 3\,\text{O}_2\text{(g)} \rightarrow 2\,\text{Al}_2\text{O}_3\text{(s)}

Passivation occurs when a reaction product forms an impermeable solid layer on an electrode, blocking further reaction.
The predicted reaction is thermodynamically favoured, but the physical barrier prevents it from continuing.
Aluminium is a key example: its surface passivates instantly in air, protecting the metal underneath.
Passivation can occur at both anodes and cathodes but is most common at anodes where oxidation products may be insoluble.

Competing Reactions: A Different, Easier Reaction Wins

A third reason a predicted reaction does not occur is that a competing reaction is easier to drive and happens instead. In any real electrolytic solution, more than one possible half-reaction may exist at each electrode. The one that actually occurs is usually the one requiring the least energy — that is, the one closest to the applied voltage.

At the particle level, imagine an electrolytic cell containing a mixture of ions. At the cathode, all the positive ions are competing to gain electrons. The ion with the most positive (or least negative) standard reduction potential wins, because it needs the least energy to be reduced. At the anode, all species capable of being oxidized are competing. The species with the least positive (or most negative) standard reduction potential as a reduction is the easiest to oxidize (it has the greatest tendency to lose electrons).

The most important competing reaction in Chemistry 30 is the oxidation of water at the anode during electrolysis. When you electrolyze a solution containing, for example, sodium sulfate (

\text{Na}_2\text{SO}_4

), you might predict that sulfate ions are oxidized at the anode. However, water is oxidized instead because the oxidation of water requires less voltage than the oxidation of sulfate. The predicted sulfate oxidation does not occur — water wins the competition.

Similarly, during the electrolysis of concentrated sodium chloride solution (brine), you might predict that sodium metal is deposited at the cathode. However, the reduction of water to produce hydrogen gas occurs instead, because water is much easier to reduce than

\text{Na}^+

ions. Knowing which species actually reacts requires comparing E° values for all species present and selecting the one that is most favourable energetically at each electrode.

2\,\text{H}_2\text{O}\text{(l)} \rightarrow \text{O}_2\text{(g)} + 4\,\text{H}^+\text{(aq)} + 4\,e^-

Multiple species in solution can compete to be oxidized or reduced at each electrode.
At the cathode, the species with the most positive E° (easiest to reduce) wins.
At the anode, the species with the most negative E° as a reduction (easiest to oxidize) wins.
Water is a very common competing species: it can be both oxidized at the anode and reduced at the cathode.
The predicted reaction may not occur if a competing reaction is energetically more favourable.

Pulling It Together: A Framework for Recognizing Failures

When you are asked whether a predicted electrochemical reaction will actually occur, work through three questions in order. First, is the applied voltage (for electrolysis) or the cell voltage (for galvanic cells) large enough to overcome both the predicted energy requirement and any known overvoltage? If not, the reaction will not occur at a useful rate — this is an overvoltage failure. Second, does the electrode material or the reaction product form a passivating layer? If yes, the reaction will stop prematurely — this is a passivation failure. Third, are other species present that are easier to oxidize or reduce? If yes, the competing reaction occurs instead — this is a competing-reaction failure.

On a diploma exam, these situations often appear as scenario-based questions. A question might describe an electrolysis setup and ask why hydrogen gas is observed at the cathode instead of the predicted metal deposit. The correct reasoning is that water is easier to reduce than the metal ion (competing reaction). Another question might describe why a predicted corrosion reaction stops — the answer is likely passivation. A question describing why no product appears even though the voltage seems sufficient points to overvoltage.

Always start with the data booklet E° values to establish the prediction, then apply whichever failure condition the scenario describes. Never skip the prediction step — you cannot recognize a failure without first knowing what was predicted.

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

Use three questions: (1) Is the voltage sufficient including overvoltage? (2) Does passivation block the reaction? (3) Is a competing reaction more favourable?
Always establish the prediction from E° values first, then evaluate why it might fail.
Diploma questions provide a scenario; your job is to match the described observation to the correct failure mechanism.
Overvoltage, passivation, and competing reactions are independent failure modes — a real situation may involve more than one.

Summary: Three Reasons Predicted Electrochemical Reactions Do Not Occur

Failure Mechanism	What Happens	Key Signal in a Question	Common Example
Overvoltage	The applied voltage is not high enough to overcome the kinetic barrier at the electrode surface, so the reaction does not start at a useful rate.	Reaction predicted to occur at a given voltage, but no product is observed until a higher voltage is applied.	Electrolysis of water requires ~1.8 V in practice, not the predicted ~1.23 V.
Passivation	The reaction produces an insoluble solid layer on the electrode that physically blocks further reaction.	A predicted corrosion or oxidation reaction slows and stops even though the reactants are still present.	Aluminium forms $\text{Al}_2\text{O}_3\text{(s)}$ on its surface, blocking further oxidation.
Competing Reaction	A different species at the same electrode is easier to oxidize or reduce, so it reacts instead of the predicted species.	A product different from the one predicted is observed; another oxidizable or reducible species is present.	Water is oxidized at the anode instead of sulfate ions in $\text{CuSO}_4\text{(aq)}$ electrolysis.

Worked example

Electrolysis of Aqueous Copper(II) Sulfate: Which Reaction Actually Occurs at the Anode?

An electrolytic cell contains aqueous copper(II) sulfate,

\text{CuSO}_4\text{(aq)}

, with inert graphite electrodes. A student predicts that sulfate ions,

\text{SO}_4^{2-}\text{(aq)}

, will be oxidized at the anode. At the cathode, the student predicts copper metal will be deposited. After running the cell, copper metal is indeed deposited at the cathode, but oxygen gas is observed at the anode — not the predicted sulfate oxidation. (a) Write the two competing half-reactions at the anode and identify which one actually occurs. (b) Use standard reduction potentials to explain why the predicted sulfate reaction does not occur. (c) Name the failure mechanism.

Step 1 — Identify the species present and possible anode half-reactions
At the anode, oxidation occurs. The species in solution that could be oxidized are water molecules and sulfate ions. Write both as reduction half-reactions first (as listed in the data booklet), then reverse them to get the oxidation half-reactions.
$\text{Reduction: } \text{O}_2\text{(g)} + 4\,\text{H}^+\text{(aq)} + 4\,e^- \rightarrow 2\,\text{H}_2\text{O}\text{(l)}, \quad E° = +1.23\,\text{V}$
Step 2 — Look up E° for sulfate oxidation
Sulfate oxidation is the reverse of the persulfate reduction half-reaction. The data booklet lists the reduction: $\text{S}_2\text{O}_8^{2-} + 2e^- \rightarrow 2\,\text{SO}_4^{2-}$ , $E° = +2.01\,\text{V}$ . Reversing this gives the oxidation of sulfate, which has a potential of $-2.01\,\text{V}$ (as an oxidation). The oxidation of water has a potential of $-1.23\,\text{V}$ (as an oxidation, i.e., the reverse of its $+1.23\,\text{V}$ reduction).
E°_{oxidation of water} = -1.23 V vs. E°_{oxidation of sulfate} = -2.01 V
Step 3 — Determine which oxidation is more energetically favourable
At the anode, the oxidation that requires the least energy wins. The oxidation requiring the least energy is the one with the least negative (or most positive) oxidation potential. Oxidation of water at $-1.23\,\text{V}$ is less negative than oxidation of sulfate at $-2.01\,\text{V}$ . Therefore, water is oxidized preferentially.
$-1.23\,\text{V} > -2.01\,\text{V}$
Step 4 — Write the actual anode half-reaction
Since water is oxidized, oxygen gas and hydrogen ions are produced at the anode. This matches the observation of oxygen gas bubbles.
$2\,\text{H}_2\text{O}\text{(l)} \rightarrow \text{O}_2\text{(g)} + 4\,\text{H}^+\text{(aq)} + 4\,e^-$
Step 5 — Name the failure mechanism
The sulfate oxidation does not occur because a competing reaction — the oxidation of water — is energetically easier (requires less applied voltage). This is a competing reaction failure. The prediction was not wrong in isolation, but the presence of water in the solution means water wins the competition at the anode.
$\Delta E°_{\text{cell, water oxidized}} = E°_{\text{cathode}} - E°_{\text{anode}} = +0.34\,\text{V} - (+1.23\,\text{V}) = -0.89\,\text{V}$

Answer: The oxidation of water (

2\,\text{H}_2\text{O}\text{(l)} \rightarrow \text{O}_2\text{(g)} + 4\,\text{H}^+\text{(aq)} + 4\,e^-

) occurs at the anode instead of sulfate oxidation. Water is easier to oxidize because its oxidation potential (

-1.23\,\text{V}

) is less negative than that of sulfate (

-2.01\,\text{V}

). The failure mechanism is a competing reaction.

Check: The cathode half-reaction is

\text{Cu}^{2+}\text{(aq)} + 2\,e^- \rightarrow \text{Cu}\text{(s)}

,

E° = +0.34\,\text{V}

. The anode half-reaction (water oxidized) has

E° = +1.23\,\text{V}

as a reduction, so the cell voltage required is

-(0.34 - 1.23) = +0.89\,\text{V}

of supplied voltage — a real, positive value that makes sense for an electrolytic cell needing an external supply. Oxygen gas at the anode and copper deposit at the cathode match the described observation, confirming the competing-reaction explanation is correct.

Common mistakes and how to avoid them

Assuming a positive

E°_{\text{cell}}

always means the reaction will definitely occur under real conditions.

Correction: A positive

E°_{\text{cell}}

predicts spontaneity under standard conditions, but passivation, overvoltage, or a competing reaction can still prevent the reaction from occurring in practice.

Confusing overvoltage with the predicted cell voltage. Students sometimes think overvoltage replaces the

E°_{\text{cell}}

calculation.

Correction: Overvoltage is an additional voltage on top of the predicted minimum. The total voltage needed is

|E°_{\text{cell}}| + V_{\text{overvoltage}}

, not just the overvoltage alone.

Forgetting that water itself can be oxidized or reduced and is therefore always a competing species in aqueous solutions.

Correction: Always include water as a possible half-reaction candidate when predicting what occurs at each electrode in an aqueous electrolytic cell.

Selecting the species with the most negative reduction E° as the one most easily reduced at the cathode.

Correction: At the cathode, the species with the most positive (least negative) E° is easiest to reduce and reacts preferentially. The most negative E° corresponds to the hardest species to reduce.

Describing passivation as the reaction simply 'stopping because it is finished' rather than because of a physical barrier.

Correction: Passivation stops a reaction early — reactants are still present and the reaction is still thermodynamically favoured. The block is a physical layer of product on the electrode surface, not exhaustion of reactants.

Lesson summary

A positive predicted cell voltage means a reaction is thermodynamically favoured, but it does not guarantee the reaction will occur under real conditions.
Overvoltage is extra voltage beyond the predicted minimum that must be applied before a reaction begins at a useful rate; it is caused by kinetic barriers at real electrode surfaces and is especially significant for gas-producing reactions.
Passivation occurs when an insoluble product coats the electrode surface and physically prevents further reaction, even though the reaction is still thermodynamically favoured.
Competing reactions occur when another species at the same electrode is easier to oxidize or reduce than the predicted species; the easier reaction wins and the predicted reaction does not occur.
Water is the most important competing species in aqueous electrolysis: it can be oxidized at the anode or reduced at the cathode instead of the predicted solute species.
To diagnose a failure, always start by establishing the prediction using E° values from the data booklet, then evaluate which failure condition — overvoltage, passivation, or competing reaction — matches the described observation.

Check your understanding

Question 1

An electrolytic cell is set up to electrolyze pure water. The predicted minimum voltage based on standard reduction potentials is 1.23 V. The student applies exactly 1.23 V but observes no gas production. What is the most likely explanation?

The cell has a short circuit, so no current flows.
The overvoltage for gas-producing reactions means more than 1.23 V is needed before gases are produced at a useful rate.
Water is passivating the electrode surface and blocking the reaction.
A competing reaction is reducing the water before it can be electrolyzed.

Show answer and explanation

The overvoltage for gas-producing reactions means more than 1.23 V is needed before gases are produced at a useful rate.

Gas-producing reactions have significant overvoltage — a kinetic barrier meaning more voltage than the thermodynamic minimum is required before bubbles appear. The predicted 1.23 V is correct in principle, but the actual reaction needs additional voltage to overcome the barrier at the real electrode surface. Passivation applies to solid-product layers, not gas evolution, and the competing-reaction concept does not apply here since only water is present.

Question 2

A student electrolyzes a solution of sodium sulfate,

\text{Na}_2\text{SO}_4\text{(aq)}

. The student predicts that

\text{Na}\text{(s)}

will be deposited at the cathode (

E° = -2.71\,\text{V}

for

\text{Na}^+/\text{Na}

). Instead, hydrogen gas is observed. Which explanation is correct?

Sodium ions are passivating the cathode surface.
The overvoltage for sodium deposition exceeds the power supply's capacity.
Water is reduced at the cathode instead of sodium ions because water has a less negative reduction potential than $\text{Na}^+$ .
Sulfate ions are being reduced at the cathode in place of sodium ions.

Show answer and explanation

Water is reduced at the cathode instead of sodium ions because water has a less negative reduction potential than

\text{Na}^+

.

At the cathode, the species with the least negative (most positive) reduction E° is reduced preferentially. Water has a reduction E° of

-0.83\,\text{V}

(in neutral solution), which is much less negative than

-2.71\,\text{V}

for

\text{Na}^+

. Water is far easier to reduce, so it wins the competition and hydrogen gas is produced. This is a competing-reaction failure. Sulfate ions are not easily reduced under these conditions, and passivation applies to solid product layers, not aqueous ion discharge.

Question 3

Aluminium metal is predicted to corrode rapidly in air based on its very negative standard reduction potential (

E° = -1.66\,\text{V}

for

\text{Al}^{3+}/\text{Al}

). Yet aluminium drink cans last for years without significant corrosion. What is the best electrochemical explanation?

The standard reduction potential for aluminium is actually positive, so corrosion is not predicted.
Aluminium forms a thin, impermeable oxide layer on its surface that passivates the metal and blocks further oxidation.
The overvoltage for aluminium oxidation is so large that the reaction cannot start.
Oxygen in air competes with aluminium and reacts with itself instead.

Show answer and explanation

Aluminium forms a thin, impermeable oxide layer on its surface that passivates the metal and blocks further oxidation.

Aluminium's corrosion is predicted and initially occurs, but the product — aluminium oxide,

\text{Al}_2\text{O}_3\text{(s)}

— forms a hard, impermeable layer directly on the metal surface. This passivating layer seals the aluminium from contact with air and water, stopping further oxidation. The reaction is still thermodynamically favoured; it is physically blocked. This is a passivation failure, not an overvoltage or competing-reaction failure.

Question 4

During the electrolysis of concentrated

\text{NaCl}\text{(aq)}

(brine), chlorine gas,

\text{Cl}_2\text{(g)}

, is produced at the anode rather than oxygen gas. The E° for the reduction of

\text{Cl}_2

is

+1.36\,\text{V}

, and the E° for the reduction of

\text{O}_2

in acidic solution is

+1.23\,\text{V}

. Based only on E° values, which species should be oxidized at the anode?

Chloride ions, $\text{Cl}^-$ , because the E° $for$ $\text{Cl}$ _2/ $\text{Cl}$ ^- is more positive, making $\text{Cl}^-$ easier to oxidize.
Water, because the E° $for$ $\text{O}$ _2/ $\text{H}$ _2 $\text{O}$ is less positive, making water easier to oxidize.
Sodium ions, because they are positive and move toward the negative electrode.
Chlorine gas, because it is already a product and would be further oxidized.

Show answer and explanation

Water, because the E°

for

\text{O}

_2/

\text{H}

_2

\text{O}

is less positive, making water easier to oxidize.

At the anode, the species that is easiest to oxidize (least energy required) reacts. Easier oxidation corresponds to a less positive (more negative) reduction E°, because you are reversing the reduction. Water has a reduction E° of

+1.23\,\text{V}

and chloride has

+1.36\,\text{V}

. Since

+1.23\,\text{V} < +1.36\,\text{V}

, water should be easier to oxidize based on E° alone. In practice, chlorine is actually produced (due to high chloride concentration and overvoltage effects on oxygen), which is itself an example of a real result differing from the simple E°-only prediction — but based purely on the E° comparison given, water is predicted to be oxidized preferentially.

Key terms

Standard reduction potential (E°): The voltage measured for a half-reaction under standard conditions (298 K, 100 kPa, 1 mol/L) when written as a reduction. Listed in the Alberta Chemistry 30 Data Booklet.
Overvoltage (overpotential): The additional voltage, beyond the thermodynamic minimum predicted by E° values, that must be applied before a real electrode reaction occurs at a useful rate. Caused by kinetic barriers at the electrode surface.
Passivation: The process by which an insoluble, tightly bonded product forms a continuous protective layer on an electrode surface, physically blocking further reaction even though the reaction is thermodynamically favoured.
Competing reaction: A reaction at the same electrode that is more energetically favourable than the predicted reaction; it occurs instead of the predicted reaction because it requires less energy to drive.
Electrolytic cell: An electrochemical cell that uses an external power source to force a non-spontaneous chemical reaction.
Cathode: The electrode where reduction (gain of electrons) occurs. In a galvanic cell it is positive; in an electrolytic cell it is negative.
Anode: The electrode where oxidation (loss of electrons) occurs. In a galvanic cell it is negative; in an electrolytic cell it is positive.
Spontaneous reaction: A reaction that proceeds on its own without a continuous external energy input. In electrochemistry, a spontaneous cell reaction has a positive $E°_{\text{cell}}$ .

Continue through CHEM30

View the complete CHEM30 Alberta Grade 12 Chemistry curriculum and lessons

About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Chemistry (CHEM30), expectation 30-B2.4k. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question

30-B2.4k · Recognize when predicted electrochemical reactions do not occur

Alberta Grade 12 Chemistry

Alberta Chemistry 30 — Unit B2: Electrochemical Cells (Outcome 30-B2.4k)

What you will learn

Prerequisite Review: How We Predict Electrochemical Reactions

Overvoltage: The Extra Push a Real Reaction Needs

Passivation: When a Protective Layer Stops the Reaction

Competing Reactions: A Different, Easier Reaction Wins

Pulling It Together: A Framework for Recognizing Failures

Summary: Three Reasons Predicted Electrochemical Reactions Do Not Occur

Electrolysis of Aqueous Copper(II) Sulfate: Which Reaction Actually Occurs at the Anode?

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through CHEM30

About this lesson and its review