Chemistry of Solutions 202-SN2-RE • CEGEP, Montreal

Revision sheet: redox and electrochemistry (202-SN2-RE)

Electrochemistry links three quantities that say the same thing in three forms: the potential $E^{\circ}$ , the free energy change $\Delta G^{\circ}$ and the equilibrium constant $K$ . Moving from one to another is half the chapter; the other half is balancing half-reactions cleanly and not confusing the two electrodes.

This sheet fixes the sign conventions, gives the three relations and their pitfalls, and gathers the mistakes that come back from one exam to the next on potentials, on Nernst and on Faraday's laws.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A standard potential is an INTENSIVE property: it is never multiplied by a stoichiometric coefficient. Doubling a half-reaction doubles the number of electrons, never the voltage, and this is the most frequent mistake of the whole chapter.

This chapter is part of 202-SN2-RE, Chemistry of Solutions

The essentials

Oxidation numbers and vocabulary

• Element in its elemental form: $0$ . Monatomic ion: its charge. $O$ is $-2$ EXCEPT in peroxides, where it is $-1$ . $H$ is $+1$ EXCEPT in metal hydrides, where it is $-1$ .
• The sum of the oxidation numbers is $0$ for a molecule and the CHARGE for a polyatomic ion.
• OXIDATION: loss of electrons, INCREASE in oxidation number, at the ANODE. The reducing agent is the one that is oxidized.
• REDUCTION: gain of electrons, DECREASE in oxidation number, at the CATHODE. The oxidizing agent is the one that is reduced.
• Anode and cathode are defined by the REACTION, never by the sign: the anode is negative in a galvanic cell and positive in an electrolytic cell.

Electrons ALWAYS flow from the anode to the cathode in the external circuit. The salt bridge does not carry electrons: it lets ions through to keep both compartments electrically neutral.

The mnemonic that does not fail: anode and oxidation both begin with a vowel, cathode and reduction with a consonant. The sign, for its part, changes with the setup.

The three ways of writing the same information

• $E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$ , both potentials being read as REDUCTION potentials from the table.
• You NEVER multiply $E^{\circ}$ by a coefficient: it is an intensive quantity, like a temperature.
• $\Delta G^{\circ}=-nFE^{\circ}$ , with $F=96\,500$ C/mol and $n$ the number of electrons EXCHANGED in the balanced overall equation.
• $\log K=\dfrac{nE^{\circ}}{0.0592}$ at $25$ °C, a direct consequence of $\Delta G^{\circ}=-RT\ln K$ .
• Spontaneous if $E^{\circ}>0$ , hence if $\Delta G^{\circ}<0$ , hence if $K>1$ . The three criteria say exactly the same thing.

The three quantities can be deduced from one another: knowing one means knowing all three. Only one of them,

E^{\circ}

, is never multiplied by a coefficient.

$E^{\circ}$ is not multiplied, but $\Delta G^{\circ}$ is: it is an extensive quantity. That is why half-reactions are combined through their $\Delta G^{\circ}$ when the coefficients do not cancel.

Nernst, electrolysis and Faraday

• Away from standard conditions: $E=E^{\circ}-\dfrac{0.0592}{n}\log Q$ at $25$ °C, with $Q$ written as for an equilibrium, products over reactants.
• At equilibrium, the cell is dead: $E=0$ and $Q=K$ . This is exactly what the relation between $E^{\circ}$ and $K$ says.
• CONCENTRATION cell: the two couples are identical, so $E^{\circ}=0$ , and all the voltage comes from the Nernst term.
• Electrolysis: $q=It$ in coulombs, then $n(e^{-})=\dfrac{q}{F}$ , then the stoichiometry of the half-reaction gives the amount of substance deposited.
• Cathodic protection: the metal to be protected is connected to a MORE REDUCING metal, that is, one with a more NEGATIVE potential, which sacrifices itself in its place.

The further to the RIGHT the potential, the stronger the oxidizing agent of the couple. In a galvanic cell, the cathode is always the couple furthest right, and the anode the one furthest left.

In an aqueous electrolysis, water itself is a candidate at both electrodes: compare the potentials before concluding that an ion discharges, or you will predict sodium deposits that never happen.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Multiplying a standard potential by a stoichiometric coefficient

the whole question, 5 points

What not to write

“We double the copper half-reaction, so $E^{\circ}$ goes from $+0.34$ to $+0.68$ V.”

What to write

“ $E^{\circ}$ is an INTENSIVE quantity: it stays $+0.34$ V. It is $n$ that doubles, hence $\Delta G^{\circ}$ , but never the voltage.”

Why: A voltage does not depend on the amount of substance, just as a temperature does not depend on the volume of water. Two identical cells in series double the voltage; doubling an equation does not.

2. Adding the two potentials instead of subtracting them

the whole question, 6 points, with an inverted conclusion

What not to write

“ $E^{\circ}_{cell}=0.34+(-0.76)=-0.42$ V, so the cell does not work.”

What to write

“ $E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=0.34-(-0.76)=+1.10$ V: the Daniell cell works perfectly.”

The two table potentials are SUBTRACTED: the gap between copper and zinc is

1.10

V. Adding them puts the cell at

-0.42

V, as if it did not work.

Why: Both table values are REDUCTION potentials. Since the anode actually undergoes oxidation, its potential must be subtracted, which amounts to changing the sign of its reversed half-reaction.

3. Identifying the anode by its sign rather than by the reaction

3 points, and half of an electrolysis question

What not to write

“The anode is always the positive terminal.”

What to write

“The anode is where OXIDATION takes place, always. It is NEGATIVE in a galvanic cell and POSITIVE in an electrolytic cell.”

In both setups, the anode is the site of OXIDATION and the electrons leave from it; only its sign changes, negative in the galvanic cell, positive in the electrolytic cell.

Why: In a galvanic cell, oxidation releases electrons that make the electrode negative. In an electrolytic cell, the external generator imposes the signs, and it draws anions toward its positive terminal, where they are oxidized.

4. Taking the number of electrons from a single half-reaction

the whole energy question, 5 points

What not to write

“The iron half-reaction exchanges $1$ electron and the permanganate one exchanges $5$ : I take $n=1$ .”

What to write

“ $n$ is the number of electrons EXCHANGED in the balanced overall equation, that is, after multiplying: here $n=5$ .”

Why: The number $n$ counts the electrons that actually cross the circuit for one mole of reaction. It is read from the overall equation, never from a half-reaction taken in isolation.

5. Writing the Nernst quotient upside down

3 points, and a voltage on the wrong side of

E^{\circ}

What not to write

“ $E=E^{\circ}-\dfrac{0.0592}{n}\log\dfrac{[\text{reactants}]}{[\text{products}]}$ .”

What to write

“ $Q$ is written like any equilibrium constant: PRODUCTS over REACTANTS. Diluting the products must RAISE the voltage, which gives the direction check.”

Daniell cell,

n=2

: with

Q=100

E=1.04

V, below

E^{\circ}

(green point). The inverted quotient puts the voltage at

1.16

V, on the wrong side of

E^{\circ}

(red point).

Why: The error is invisible under standard conditions, since $Q=1$ and $\log Q=0$ . It shows up as soon as a concentration departs from unity, and physical sense settles it immediately.

6. Giving $-2$ to oxygen in a peroxide

3 points, and a half-reaction impossible to balance

What not to write

“In $H_{2}O_{2}$ , oxygen is $-2$ , so hydrogen is $+2$ .”

What to write

“In a PEROXIDE, oxygen is $-1$ : $2(+1)+2(-1)=0$ , which is consistent. Hydrogen stays at $+1$ .”

In the peroxide, the two oxygens are bonded to each other: each is

-1

, and hydrogen stays at

+1

as in water.

Why: An oxidation number of $+2$ for hydrogen is impossible, since it has only one electron. That check flags the exception without having to know it by heart.

7. Balancing directly with hydroxide ions in basic solution

the whole balancing question, 5 points

What not to write

“The solution is basic, so I balance the hydrogens with $OH^{-}$ right from the start.”

What to write

“We ALWAYS balance in acidic solution first, with water and $H^{+}$ ; then we add as many $OH^{-}$ as $H^{+}$ on BOTH SIDES and cancel into water.”

Why: The two-step method is mechanical and does not go wrong. Balancing directly in basic solution means guessing how much water to add on each side, which nobody gets right under pressure.

8. Believing a concentration cell cannot deliver any voltage

the whole question, 5 points

What not to write

“Both compartments contain the same couple, so $E^{\circ}=0$ and the cell is dead.”

What to write

“ $E^{\circ}=0$ but $E\neq 0$ : all the voltage comes from the Nernst term, $E=-\dfrac{0.0592}{n}\log\dfrac{[\text{dilute}]}{[\text{concentrated}]}$ .”

Same couple on both sides, so

E^{\circ}=0

; the voltage comes from the concentration difference. The dilute side is the anode: copper dissolves there and tends to equalize the two compartments.

Why: The system tends to equalize the two concentrations, and this spontaneous transfer is what supplies the electrical work. The cell stops exactly when the two compartments are identical.

9. Forgetting the electron stoichiometry in a Faraday's law calculation

3 points, and a mass off by a factor of two

What not to write

“ $0.0746$ mol of electrons flowed, so $0.0746$ mol of copper was deposited.”

What to write

“ $Cu^{2+}+2e^{-}\rightarrow Cu$ : TWO electrons are needed per atom, so only $0.0373$ mol of copper.”

Why: The quantity of electricity gives moles of ELECTRONS, never directly moles of metal. The conversion between the two is read from the half-reaction, like any stoichiometry.

10. Protecting a metal with a less reducing metal

4 points on the reasoning question

What not to write

“To protect an iron hull, we connect it to copper, which is more noble.”

What to write

“We need a MORE REDUCING metal, with a more NEGATIVE potential, such as zinc or magnesium: it is the one that is oxidized in place of the iron. Copper would do exactly the opposite and speed up corrosion.”

The metal with the LOWEST potential is the anode and dissolves: connected to zinc, iron is protected; connected to copper, it is the one that corrodes.

Why: In a pair of metals in contact, the one with the lowest potential plays the role of anode and dissolves. Connecting iron to copper makes iron the anode, hence the sacrificed piece.

Which method to choose

Which relation to use

The quantity given and the quantity asked for

If two table potentials, and the voltage is asked for → $E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$ , never multiplying

Example: $0.34-(-0.76)=1.10$ V
If the maximum energy or a quantified spontaneity is asked for → $\Delta G^{\circ}=-nFE^{\circ}$ , with $n$ read from the OVERALL equation

Example: $-2\times 96\,500\times 1.10=-212$ kJ/mol
If the equilibrium constant is asked for → $\log K=\dfrac{nE^{\circ}}{0.0592}$

Example: $\log K=37.2$ , so $K\approx 1.6\times 10^{37}$
If the concentrations are not $1$ mol/L → Nernst, $E=E^{\circ}-\dfrac{0.0592}{n}\log Q$

Example: $1.10-0.0296\log 100=1.04$ V
If a current and a duration are given → Faraday: $q=It$ , $n(e^{-})=\dfrac{q}{F}$ , then the stoichiometry of the half-reaction

the unit of $q$ is the coulomb, so the time must be in SECONDS

The first three branches give three expressions of the same thing. A spontaneous reaction has $E^{\circ}>0$ , $\Delta G^{\circ}<0$ and $K>1$ simultaneously: if the three do not agree, there is a sign error.

Balancing a half-reaction

The medium stated in the problem, and the order of operations

If ACIDIC solution → balance the element, then $O$ with $H_{2}O$ , then $H$ with $H^{+}$ , then the CHARGE with electrons

Example: $MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O$
If BASIC solution → do all the work in acidic solution first, THEN add as many $OH^{-}$ as $H^{+}$ to both sides and cancel

Example: the $H^{+}$ become water, the water cancels
If the two half-reactions must be combined → multiply each one to equalize the number of electrons, then add, cancelling the electrons

Example: $5$ times the iron against $1$ time the permanganate
If you doubt the spontaneous direction → compare the two potentials: the couple with the HIGHER potential is reduced

Example: copper is reduced, zinc is oxidized
If you do not know which element changes oxidation state → calculate the oxidation numbers of all the elements before and after

an element whose oxidation number does not change is a spectator and appears in no half-reaction

The order element, oxygen, hydrogen, charge is mechanical and not up for debate. At the end, check that the total charge AND the number of each atom are the same on both sides: two independent checks.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Describing a galvanic cell completely

When to use it: The problem gives two couples and asks for the voltage, the equation, the spontaneity or the constant.

1 Read the two REDUCTION potentials from the table, and designate as CATHODE the couple with the higher potential.
2 Write the two half-reactions in the right direction: reduction at the cathode, oxidation at the anode, the latter being the reverse of the table line.
3 Multiply the half-reactions to equalize the electrons, add, and read $n$ from the overall equation.
4 Calculate $E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$ , WITHOUT multiplying the potentials.
5 Deduce $\Delta G^{\circ}=-nFE^{\circ}$ and, if the problem asks for it, $\log K=\dfrac{nE^{\circ}}{0.0592}$ .
6 Conclude on the spontaneity and check that the three criteria agree.

Concluding sentence

“The copper couple has the higher potential, so it is the cathode: $E^{\circ}_{cell}=0.34-(-0.76)=1.10$ V, with $n=2$ , hence $\Delta G^{\circ}=-212$ kJ/mol and $\log K=37.2$ : the reaction is overwhelmingly spontaneous.”

The trap: Multiplying the half-reactions AND the potentials. Only the half-reactions are multiplied, because we are equalizing electrons; the potentials do not depend on the amount of substance.

Marking: Typically 1 point for identifying the electrodes, 2 for the half-reactions, 1 for the overall equation and $n$ , 2 for the voltage, 2 for the free energy and the constant.

Handling an electrolysis with Faraday's laws

When to use it: The problem gives a current and a duration, and asks for a mass deposited, a volume of gas or a time required.

1 Write the half-reaction that takes place at the electrode in question, and count the electrons it exchanges.
2 Convert the duration to SECONDS, then calculate $q=It$ in coulombs.
3 Calculate $n(e^{-})=\dfrac{q}{F}$ with $F=96\,500$ C/mol.
4 Divide by the number of electrons in the half-reaction to get the amount of substance formed, then convert to mass or volume.
5 Check the order of magnitude: one ampere for one hour deposits about a gram of copper, a little more or less depending on the metal.

Concluding sentence

“A current of $1.00$ A for $2.00$ h supplies $q=7200$ C, that is $0.0746$ mol of electrons; since two are needed per copper atom, $0.0373$ mol is deposited, that is, $2.37$ g.”

The trap: Leaving the duration in hours. The coulomb is an ampere-second, and forgetting the conversion gives a result three thousand six hundred times too small.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The agreement of the three criteria

Check that $E^{\circ}>0$ , $\Delta G^{\circ}<0$ and $K>1$ go together. A disagreement between any two of them signals a sign error.

An $E^{\circ}=+1.10$ V with a positive $\Delta G^{\circ}$ : the minus sign of the relation was dropped.
The potential never multiplied

Go back over the calculation and check that no coefficient touched an $E^{\circ}$ . Only $n$ and the half-reactions are multiplied.

An $E^{\circ}=+0.68$ V for the copper couple: the half-reaction was doubled and the potential with it.
Charge and atoms on both sides

After balancing, count each element AND the total charge on both sides. Two independent checks that take ten seconds.

An equation whose charge is $+2$ on the left and $+7$ on the right: the electrons were not added.
The direction of the Nernst shift

Diluting the PRODUCTS must raise the voltage, diluting the REACTANTS must lower it. Check that the result goes in that direction.

A voltage above $E^{\circ}$ when a reactant was diluted: the quotient is written upside down.
The order of magnitude of an electrolysis

One ampere for one hour passes about $0.037$ mol of electrons, so a few grams of metal at most. A result in kilograms is wrong.

A deposit of $8.5$ kg for two hours at one ampere: the time, already in seconds, was multiplied by $3600$ a second time.

The typical problem, taken apart

The Daniell cell, from the voltage to the mass of zinc consumed

We build a cell by dipping a zinc electrode into a zinc sulfate solution and a copper electrode into a copper sulfate solution, the two compartments being joined by a salt bridge.

Determine the overall equation, the standard voltage, the standard free energy change and the equilibrium constant. Then calculate the actual voltage if $[Zn^{2+}]=1.00$ mol/L and $[Cu^{2+}]=0.010$ mol/L, then the mass of zinc consumed if the cell delivers $1.00$ A for $2.00$ h.

Data: $E^{\circ}(Cu^{2+}/Cu)=+0.34$ V, $E^{\circ}(Zn^{2+}/Zn)=-0.76$ V, $F=96\,500$ C/mol, $M(Zn)=65.38$ g/mol.

The cell in the problem: zinc in

Zn^{2+}

1.00

mol/L, copper in

Cu^{2+}

0.010

mol/L, salt bridge and voltmeter.

Step 1

The copper couple has the higher potential: it is REDUCED, so copper is the CATHODE. Zinc is oxidized, it is the ANODE. Half-reactions: $Cu^{2+}+2e^{-}\rightarrow Cu$ and $Zn\rightarrow Zn^{2+}+2e^{-}$ .

Why

The electrodes are chosen by comparing potentials, never by habit. The anode half-reaction is the REVERSE of the table line, and writing it out avoids the next sign error.

Step 2

Overall equation: $Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$ , with $n=2$ electrons exchanged.

Why

Both half-reactions already exchange two electrons: no multiplication is needed, and $n$ is read directly from the overall equation.

Step 3

$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=0.34-(-0.76)=+1.10$ V.

Why

Both values are reduction potentials: the subtraction reflects the fact that the anode runs in reverse. No coefficient comes into play here, even if the equation carried some.

Step 4

$\Delta G^{\circ}=-nFE^{\circ}=-2\times 96\,500\times 1.10=-2.12\times 10^{5}$ J/mol, that is $-212$ kJ/mol.

Why

This is where $n$ comes in, and it alone carries the multiplication. Free energy is extensive: doubling the equation would double it.

Step 5

$\log K=\dfrac{nE^{\circ}}{0.0592}=\dfrac{2\times 1.10}{0.0592}=37.2$ , so $K\approx 1.6\times 10^{37}$ .

Why

A constant of this order means the reaction is practically complete: no measurable copper ions remain when the cell stops.

Step 6

Nernst: $Q=\dfrac{[Zn^{2+}]}{[Cu^{2+}]}=\dfrac{1.00}{0.010}=100$ , so $E=1.10-\dfrac{0.0592}{2}\log 100=1.10-0.059=1.04$ V.

Why

The quotient is written products over reactants, like any equilibrium constant. Diluting the reactant $Cu^{2+}$ makes the voltage DROP, which is the expected direction check.

Step 7

Faraday: $q=It=1.00\times 7200=7200$ C, so $n(e^{-})=\dfrac{7200}{96\,500}=0.0746$ mol.

Why

The duration is converted to seconds on this very line. Since the coulomb is one ampere-second, leaving the hours in would produce an error of a factor of $3600$ .

Step 8

The zinc half-reaction exchanges TWO electrons per atom: $n(Zn)=\dfrac{0.0746}{2}=0.0373$ mol, so $m=0.0373\times 65.38=2.44$ g.

Why

Moles of electrons are not moles of metal: the conversion is read from the half-reaction, exactly like ordinary stoichiometry.

Step 9

Checks: $E^{\circ}>0$ , $\Delta G^{\circ}<0$ and $K\gg 1$ agree; the actual voltage $1.04$ V is indeed BELOW $1.10$ V since the reactant was depleted.

Why

Three independent checks, all structural: they catch a lost sign or an inverted quotient without redoing a single calculation.

The conclusion, written out

“The Daniell cell delivers $1.10$ V under standard conditions, which corresponds to $\Delta G^{\circ}=-212$ kJ/mol and an equilibrium constant of the order of $10^{37}$ . With copper ions a hundred times less concentrated than zinc ions, the voltage falls to $1.04$ V, and a current of $1.00$ A for two hours consumes $2.44$ g of zinc.”

The classic mistake on this problem: Adding the two potentials and finding $-0.42$ V, then concluding that the cell does not work. The Daniell cell is the canonical example of a cell that does work: it is the sign that gives the error away, not the value.

Learn by heart

• Anode: OXIDATION, always. Negative in a galvanic cell, POSITIVE in an electrolytic cell.
• $E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$ , and $E^{\circ}$ is NEVER multiplied.
• $\Delta G^{\circ}=-nFE^{\circ}$ with $F=96\,500$ C/mol and $n$ read from the OVERALL equation.
• $\log K=\dfrac{nE^{\circ}}{0.0592}$ at $25$ °C. Spontaneous: $E^{\circ}>0$ , $\Delta G^{\circ}<0$ , $K>1$ .
• Nernst: $E=E^{\circ}-\dfrac{0.0592}{n}\log Q$ , with $Q$ as PRODUCTS over REACTANTS.
• Concentration cell: $E^{\circ}=0$ , all the voltage comes from the Nernst term.
• Faraday: $q=It$ in SECONDS, $n(e^{-})=\dfrac{q}{F}$ , then divide by the electrons of the half-reaction.
• $O$ is $-2$ except in peroxides ( $-1$ ); $H$ is $+1$ except in hydrides ( $-1$ ).

Frequently asked questions

Do you multiply the standard potential when you multiply a half-reaction?

No, never. Potential is an intensive quantity, like a temperature: it does not depend on the amount of substance. What gets multiplied is the number of electrons exchanged, and therefore the free energy change, which is extensive. Doubling an equation doubles the available energy, not the voltage.

Is the anode positive or negative?

It depends on the setup. The anode is always the electrode where oxidation takes place, and that is its only definition. In a galvanic cell it is negative because oxidation releases electrons there. In an electrolytic cell, the generator imposes the signs and the anode becomes positive. So reason from the reaction, never from the sign.

How do you know which couple is reduced in a galvanic cell?

The one with the higher standard reduction potential. It forms the cathode, while the other couple runs in reverse and serves as the anode. The cell voltage is then the difference between the cathode potential and the anode potential, and it is necessarily positive if you chose correctly.

What is the value of n in the relation between free energy and potential?

It is the number of electrons actually exchanged in the balanced overall equation, after multiplying the half-reactions so that they match. It is not the number of electrons of a single half-reaction taken in isolation. A mistake on n corrupts both the free energy change and the equilibrium constant.

Can a concentration cell deliver current?

Yes, even though its standard potential is zero. The two compartments contain the same couple at different concentrations, and the system spontaneously tends to equalize them. All the voltage then comes from the Nernst term, and the cell stops exactly when the two concentrations have become identical.

How do you protect an iron object from corrosion?

By connecting it to a more reducing metal, that is, one with a more negative potential, such as zinc or magnesium. That metal becomes the anode and dissolves in place of the iron: this is called a sacrificial anode. Connecting iron to a more noble metal, such as copper, would have the opposite effect and speed up its corrosion.

Practise it

Corrected exercises: Redox and electrochemistry, 202-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Solubility and the solubility product Next sheet Chemical thermodynamics

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn2-redox-electrochemistry. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a 202-SN2-RE tutor in Montreal?

Get in touch for a first session. In electrochemistry, we first set the conventions: anode, cathode, direction of the electrons, sign of the potential. The calculations then follow on their own.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: redox and electrochemistry (202-SN2-RE)

The essentials

Oxidation numbers and vocabulary

The three ways of writing the same information

Nernst, electrolysis and Faraday

The mistakes that cost marks

1. Multiplying a standard potential by a stoichiometric coefficient

2. Adding the two potentials instead of subtracting them

3. Identifying the anode by its sign rather than by the reaction

4. Taking the number of electrons from a single half-reaction

5. Writing the Nernst quotient upside down

6. Giving −2-2−2 to oxygen in a peroxide

7. Balancing directly with hydroxide ions in basic solution

8. Believing a concentration cell cannot deliver any voltage

9. Forgetting the electron stoichiometry in a Faraday's law calculation

10. Protecting a metal with a less reducing metal

Which method to choose

Which relation to use

Balancing a half-reaction

How the answer is expected to be written

Describing a galvanic cell completely

Handling an electrolysis with Faraday's laws

Check before you hand in

The typical problem, taken apart

The Daniell cell, from the voltage to the mass of zinc consumed

Learn by heart

Frequently asked questions

Do you multiply the standard potential when you multiply a half-reaction?

Is the anode positive or negative?

How do you know which couple is reduced in a galvanic cell?

What is the value of n in the relation between free energy and potential?

Can a concentration cell deliver current?

How do you protect an iron object from corrosion?

Corrected exercises: Redox and electrochemistry, 202-SN2-RE

See also

Looking for a 202-SN2-RE tutor in Montreal?

6. Giving $-2$ to oxygen in a peroxide