Category Archives: Chem 2


Galvanic (or voltaic) cells:

A voltaic cell consists of two half cells, which are connected electrically. A half cell refers to the part of the cell in which one half reaction occurs, and in all text books is depicted as a metal electrode sitting in a beaker of electrolyte solution.

Yeah, you all know this diagram


All voltaic cells share several features, which are essential for operation:

Half cells:

The half cell must contain an electrode, (if the reaction involves reduction to or oxidisation of the metal, the electrode will generally be that metal. If neither species in a half reaction is metal, an inert electrode such as platinum will be used). The solution must contain the relevant ions for the redox reaction to occur.

External circuit:

The half cells need to be connected via a wire, so electrons can be transferred from one half cell to the other. The circuit may be connected via a voltmeter, so the potential of the cell can be measured.

Salt bridge:

If a voltaic cell runs without a salt bridge, charge will quickly build up within the half cells and the cell would stop working. A salt bridge allows ions to flow into each half cell, balancing the charge and allowing extended operation of the cell, but preventing mixing of the two solutions.

Drawing galvanic cells is fairly straightforward – the skeleton two-beaker sketch is a good starting point. We’ll go over how to draw one using an example:

Zn(s) + Cu2+(aq)  →  Cu(s) + Zn2+(aq)

The first thing to do is break it up into the half reactions and identify which is oxidation and which is reduction:

Zn(s) → Zn2+(aq) + 2e                                                                oxidation

Cu2+(aq) + 2e → Cu(s)                                                               reduction

The anode is the half cell in which oxidation occurs, and reduction occurs in the cathode (an ox and a red cat). The anode is always depicted as being on the left, the cathode the right. Label the anode and the cathode, showing what metals and ions are involved:

Show the direction of electron flow (always anode to cathode) and ions in the salt bridge:

This is a fairly clumsy way of depicting cells, and there is a shortcut known as the “line notation”.

Line notation.

As with sketching a cell, the anode is always on the left and the two half cells are separated by a double line, eg:


 Single lines indicate a change in state. The electrodes are always shown on the outer edges, and the reactants are shown in the same order as they would be written in the half reaction. If the half cell includes a gaseous, or two aqueous reactants, an inert electrode is used and any reactants in the same state are separated by a comma, eg:

 Zn(s)|Zn2+(aq)||Fe3+(aq), Fe2+(aq)|Pt

 Cell potentials:

The cell potential tells you how much of a driving force there is behind the cell reaction, and is the difference between the two half-cell potentials.

Cell potential = oxidation potential + reduction potential

Or… because your potentials are usually given as reduction potentials:

Cell potential = reduction potential of cathode – reduction potential of anode

 A note about potentials:

The more positive the standard potential is, the better the species is at oxidising. The more negative, the better the reductant.

Nernst Equation:

The cell potential is easy to work out if you are working in standard conditions, but will change if the concentrations of the components change. Luckily for you, Nernst worked it all out for you and you just need to remember his equation:

Keeping in mind that Q is the reaction quotient and essentially the equilibrium constant (concentration of products/concentration of reactants) when the system isn’t at equilibrium. The reaction is the same for the equilibrium constant, except that at equilibrium the Ecell will be zero.


Some links:


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Chem 2 tute sheet week 6 – Electrochemistry

  1. Zinc reacts spontaneously with the silver ion:

Zn­(s) + 2Ag+(aq) →  Zn2+(aq) + 2Ag(s)

Describe a voltaic cell using this reaction. What are the half reactions?

  1. Consider the cell :


Write the half cell reactions and the overall cell reaction. Make a sketch of this cell and label it. Include labels showing the anode, cathode, and direction of the electron flow.


Standard reduction potentials are 2.9 V for F2/F-, 0.8 V for Ag+/Ag, 0.5 V for Cu+/Cu, 0.3 V for Cu2+/Cu, -0.4 V for Fe2+/Fe, -2.7 V for Na+/Na, and -2.9 V for K+/K.

a. Arrange the oxidising agents in order of increasing strength.

b. Which of these oxidising agents will oxidize Cu under standard state conditions?

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Oxidation and reduction or, all the things you have forgotten from high school but are expected knowledge at university.

Oxidation/reduction reactions involve the transfer of electrons from one species to another, oxidising one species and reducing the other.

Zn(s) + Cu2+(aq) →  Zn2+(aq) + Cu(s)

In this classic example, the zinc metal gives up two electrons and becomes the zinc (II) ion and the copper (II) ions gain two electrons to become copper metal. The transfer of electrons in redox reactions can be kept track of using oxidation numbers.

Oxidation numbers:

An oxidation number (or state) of a species gives you the hypothetical charge of an atom in a substance, (or actual charge, if it is a monatomic ion) as determined by a few rules. The oxidation numbers will change during an oxidation/reduction reaction, as electrons will be transferred.

The rules for determining the oxidation state are as follows:

  1. The oxidation state of an element (eg. Fe(s), Ca(s), Ar(g)) is always zero.
  2. The oxidation state of an atom in a monatomic ion is the same as the charge on the ion.
  3. The oxidation state of oxygen is always -2, (unless it is a peroxide, in which case it is -1)
  4. The oxidation state of hydrogen is usually +1 (unless it is in a compound such as CaH2, where it is -1)
  5. The oxidation state of fluorine is always -1. Other halogens are usually -1, except when one of the above rules takes precedence.
  6. The sum of the oxidation numbers in a neutral compound is zero. In an ion, the sum of the oxidation numbers is equal to the charge on the ion.

Half reactions:

Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)

Shown above is the overall equation for the oxidation of iron with copper (II), but we can also write this reaction in terms of two half reactions. A half reaction is one part of the oxidation/reduction reaction, showing what happens to each species separately. The reaction shown above can be split accordingly:

Fe(s) → Fe2+(aq) + 2e                            electrons lost by Fe

Cu2+(aq) + 2e → Cu(s)                           electrons gained by Cu

The oxidation reaction is the one in which there is a loss of electrons and increase in the oxidation number, (in this case, the iron). The reduction is the gain of electrons and decrease in oxidation number, (the copper). Confusingly, we refer to each species by what it does, not what happens to it. So the oxidising agent, (or oxidant) is the species that causes oxidation in something else but is itself reduced. The reductant is oxidised, but causes reduction in something else. Don’t get this confused.

You may not always be given the full half equation, you may simply be told the redox couple, (eg, Fe/Fe2+, MnO4/Mn2+) and you will have to balance it yourself. Balancing a half equation is fairly simple, you just need to follow the steps:

  1. Assign oxidation numbers so you know what is oxidised/reduced.
  2. Balance the half equation by:
    1. Balance all the atoms except oxygen and hydrogen
    2. Balance the oxygens by adding the appropriate number of water molecules to the other side of the equation.
    3. Balance the hydrogen atoms, (including the ones you have added in the form of water) by adding H+.
    4. Balance the charge by either i) adding the appropriate number of electrons that have been lost/gained as seen by the change in oxidation number or ii) add electrons where necessary to make sure the equation is neutral overall.
  3. Once you’ve done that for both half equations, add them together. You may need to multiply one or both by a factor to make sure that the electrons have cancelled. If you still have electrons in your overall equation, you’ve done something wrong.

Let’s try an example (adapted from Ebbing General Chemistry):

 Zn(s) + NO3(aq) → Zn2+(aq) + NH4+(aq)

 We can split this one into two half equations:

Zn(s) → Zn2+(aq)­

NO3(aq) → NH4+(aq)

The zinc reaction is easy – the oxidation state starts at zero and ends up at +2, so it is an oxidation requiring two electrons:

Zn(s) → Zn2+(aq)­ + 2e

 The nitrogen half reaction is more complex. First, we need to determine the oxidation states of nitrogen:


The oxygens must each have an oxidation number of -2, and there is an overall -1 charge. Therefore the nitrogen must be +5.


The hydrogens are each +1, and there is an overall +1 charge, so nitrogen must be -3 in this case.

NO3(aq) → NH4+(aq)

Now we need to balance the equation – the number of nitrogens is correct, so we move to the oxygens:

NO3(aq) → NH4+(aq) + 3H2O(l)

Now we need to balance the hydrogens, including those on the ammonium:

NO3(aq) + 10H+(aq) → NH4+(aq) + H2O(l)

Finally, the charge:

NO3(aq) + 10H+(aq) + 8e NH4+(aq) + H2O(l)

We now need to multiply the zinc half reaction by 4, so a total of 8 electrons are being transferred in both reactions, and add them together to give:

4Zn(s) + NO3(aq) + 10H+(aq) → 4Zn2+(aq) + NH4+(aq) + 3H2O(l)

These steps work just fine for acidic reactions, but if your reaction takes place under basic conditions, you need to balance all of the protons with hydroxide ions.

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Chem 2 tute sheet week 5 (22/08)

Give the major product (or products) that would be obtained when each of the following compounds is reacted with Cl2 and AlCl3:

  • Ethylbenzene
  • Nitrobenzene
  • Chlorobenzene
  • Benzoic acid

Starting with benzene, outline a synthesis of each of the following:

  • Isopropylbenzene
  • Nitrobenzene
  • Tert-butylbenzene
  • m-dinitrobenzene
  •  m-bromonitrobenzene
  • p-bromonitrobenzene

Show the mechanisms for the following:

  • Addition of HBr to ethanal
  • Substitution of HBr and acetyl chloride (CH3COCl)
  • Why is a) an addition and b) a substitution?

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Chem 2 tute sheet for week 4 (15/08)

  1.  What is Markovnikov’s law? Why do reactions proceed according to it?
  2. Give the product (or products) that you would expect to be formed in each of the following reactions. In each case, give the mechanism, (Sn1, Sn2, E1 or E2) by which the product is formed and predict the relative amount of each.
    1. CH3CH2CH2Br + CH3O →
    2. CH3CH2CH2Br + (CH3)3CO– →
    3. (CH3CH2)3CBr + OH-→

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Electrophilic Aromatic Substitution

Benzene rings are rich in electrons, which exist in an electron cloud over the conjugated ring. Being so electron rich, they are attractive to electrophilic reagents. There are five main types of EAS, which are listed below:




Friedel-Crafts Alkylation

Friedel-Crafts Acylation

All of these processes occur via the same mechanism, but you will need to know what reagents and solvents are required for each reaction.

The general mechanism is as follows:

General mechanism for EAS reactions

Halogenation, (for the purposes of this example, bromination) will not occur if benzene and Br2 are mixed together, the reaction requires the presence of a Lewis acid.

FeCl3 or AlCl3 are used for chlorinations, and they follow the same procedure. Fluorination of benzene is difficult as fluorine reacts rapidly with benzene and requires special equipment and an indirect method of reaction. You don’t need to know that reaction mechanism. Iodine is very unreactive towards benzene and so reaction has to take place in the presence of an oxidising agent, such as nitric acid.

Nitration of benzene can occur by mixing hot nitric acid with benzene, but the reaction is very slow. It can be sped up by adding sulphuric acid, which increases the concentration of the electrophile, (NO2+).

Sulphonation of benzene requires “fuming sulphuric acid”, which is sulphuric acid with extra SO3 added. The sulphuric acid produces even more SO3, which is the electrophile in this reaction.

Friedel-Crafts alkylations, (named after the chemists who discovered this method, Charles Friedel and James Crafts) are valuable reactions as they form a carbon-carbon bond – a very important thing in organic syntheses. The first step in connecting an alkyl group to a benzene ring is forming a carbocation, here achieved by reacting an alkyl chloride with AlCl3. The reaction then proceeds as normal.



If the R-X is a primary alkyl chloride, (meaning that it will form an unstable primary carbocation when the chloride is removed) the AlCl3 will instead form a complex with the alkyl group, which behaves as if it were a carbocation.


Friedel-Crafts acylation is much the same as an alkylation, requiring an acid chloride and Lewis acid to react, although it will also react with acetic anhydride, (remember back to first semester and the preparation of paracetamol).

Effect of substituents on reactivity and substituent direction:


Benzene’s reactivity is can be affected by any substituents attached, and this will determine the site of attack if any more substituents are added. If a substituent makes the benzene more reactive, it is referred to as an activating group, and it if makes it less reactive, a deactivating group. So far, so good.




Activating groups are anything that will donate electrons into the ring, increasing the electron density of the benzene and making it more attractive to electrophiles. These groups increase the rate of EAS reaction. You are looking for any group with lone pairs, or alkyl groups:

Electron withdrawing groups will deactivate the benzene and generally have a partial or full positive charge adjacent to the ring or electonegative atoms to remove electron density from the benzene ring.

Activating groups are orth-para directing, as any additional substituents will be directed to the ortho or para positions due to the resonance effect of the electron donating group, which can form a relatively stable fourth resonance form.

Deactivating groups are meta-directing, as ortho/para substituents lead to unstable resonance forms with the positive charge on the ring directly next to the partial or full positive charge on the substituent.

The exception to this rule are the halogens, despite being electron withdrawing due to strong electronegativity, they still have lone pairs, which can be donated into the ring, making them ortho/para directing.


Some links: (pdf link)








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Filed under Chem 2, Organic Chemistry

Chem 2 tute sheet for week 3 (11/08)

This is a copy of the sheet I handed out in the week 3 tute:

1.  Identify the substrate, nucleophile and the leaving group:

a.    CH3CH2Br + CH3S → CH3CH2SCH3 + Br

b.   (CH3)2CHCH2OSO2F + NH3 → (CH3)2CHCH2NH3+ + FSO3

2. Show the major products for the addition reactions of a) propene and b) methylcyclohexene with the following:

i) H2/Pd

ii) HBr

iii) Br2

iv) H2O/H2SO4

3. Indicate the type of reaction shown below:

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Filed under Chem 2, Organic Chemistry, Tute sheet

Addition reactions

One example of nucleophilic/electrophilic reactions, (known as polar reactions) are additions to double bonds.

Alkenes are electron rich, so they behave as nucleophiles in polar reactions and attack electrophiles.

The placement of the new substituents is important and dictated by Markovnikov’s rule.

Vladimir Markovnikov: one of the two Russians we'll talk about during this course. This is the one with the better beard.

Markovnikov’s rule, according to McMurry’s Organic Chemistry:

In the addition of HX to an alkene, the H attaches to the carbon with fewer alkyl substituents, (the one with more hydrogens) and the X attaches to the carbon with more alkyl substituents.

In other words: the rich get richer.

The reason for this is pretty simple. Addition reactions occur in two steps, via a carbocation intermediate. The reaction won’t proceed unless the carbocation is as stable as possible, requiring the charged carbon to be the one with the most substituents.

Some further links and questions:

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Filed under Chem 2, Organic Chemistry

Nucleophiles, Electrophiles and Leaving Groups: A Spotter’s Guide

Nucleophiles and electrophiles are important classes of molecules as they are involved with a large number of organic reactions. Success and happiness in organic chemistry relies on you being able to identify which is which.


Nucleophile means “nucleus lover” and is pretty much what it says on the tin – these are c0mpounds that are strongly attracted to positive charge. It follows that they themselves are electron rich, sometimes with a formal negative charge.  Good nucleophiles are also very reactive. You know of another class of compounds that are reactive and electron rich – bases. Nucleophiles are generally strong bases,  or the conjugate base of a weak acid.


Electrophiles are the counterpart to nucleophiles. They crave electrons as they are electron poor, (sometimes positively charged). You won’t see the term “electrophile” as much as you will “nucleophile”, but you will see “electrophilic site”, which refers to the reactive, electron poor part of the substrate.

Leaving Groups:

Leaving groups are the part of the molecule that is booted off during a substitution or elimination reaction. They must be stable on their own, and not too tightly bound to the substrate. They are generally negatively charged, as they take the electrons with them when they leave. A good leaving group is a weak base, or a conjugate base of a strong acid. Note the difference between a g0od nucleophile and a good leaving group. This is an important distinction.

See? I told you, you need to know about acids and bases!

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Filed under Chem 2, Organic Chemistry