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Recognising redox reactions


Oxidation and reduction


Oxidation states

Oxidation states, sometimes called oxidation numbers, are a simple way of keeping track of redox reactions, so that it is easy to see which species has been oxidized and which reduced.

sodium atom sodium ion

+

oxidation state = 0

oxidation state = +1

The oxidation state of an element in a compound is equivalent to the number of electrons it has lost or gained by forming bonds.

They also help in balancing equations for redox reactions.


Assigning oxidation states


Assigning oxidation states

Some elements have the same oxidation state in all their compounds, or in most with some specific exceptions. It is useful to memorise the following examples:

+1 except in hydrides of group I/II metals, e.g. NaH, where it is -1

-2 except in peroxides (-1) and in compounds with fluorine (+2)

-1 except in compounds with F and O where it has positive values

-1

hydrogen

oxygen

fluorine

chlorine

Element Oxidation state in compounds


What is the oxidation state?


Redox reactions a summary



Half equations

Chloride ions can be oxidised to produce chlorine. The half equation for this reaction is:

All redox reactions can be illustrated using half equations. Half equations can be combined to give the equation for the overall redox process.

Half equations are used to show the loss or gain of electrons when a species undergoes oxidation or reduction.

2Cl– Cl2 + 2e- -1 0

One element in a half equation changes oxidation state. Here chlorine has changed its oxidation state from -1 to 0.


Combining half equations

To combine half equations:

Step 1: Write the half equations. (You may need to work these out if complex ions and other species such as H+ are involved.)

Step 2: Make sure that the number of electrons in each half equation is the same, so that the electrons cancel out. Do this by multiplying one or both equations to make the number of electrons the same in each case.

Step 3: Add the half equations and cancel the electrons. It may be possible to cancel other species that appear on both sides – often H+ or H2O.


Combining half equations – example

Chlorine oxidizes iron(II) to iron(III) and is itself reduced to chloride ions. Write a balanced equation for this reaction.

Step 3: Add the half equations and cancel the electrons.

Cl2(aq) + 2Fe2+(aq) 2Cl-

(aq) + 2Fe3+(aq)

2Fe2+(aq) 2Fe3+

(aq) + 2e-

Step 2: Eqn. A involves 2 electrons and Eqn. B involves 1 electron, so multiply both sides of Eqn. B by two.

chlorine has been reduced

Step 1: Write the half equations.

Eqn. A

Eqn. B

Cl2(aq) + 2e- 2Cl-(aq)

Fe2+(aq) Fe3+

(aq) + e- iron(II) has been oxidized


Combining half equations



What is a half cell?

If a rod of metal is dipped into a solution of its own ions, an equilibrium is set up. For example:

Zn(s) Zn2+(aq) + 2e-

This is a half cell and the strip of metal is an electrode. The position of the equilibrium determines the potential difference between the metal strip and the solution of metal.

zinc sulfatesolution

(1 mol dm-3)

zinc metalstrip


Cells and electrode potentials


Combining half cells 1


The standard hydrogen electrode


Combining half cells 2


Representing half cells: cell diagrams

An electrochemical cell can be represented in a shorthand way by a cell diagram.

The double vertical lines represents a salt bridge. The single lines represent a phase change between the solid metal and the aqueous metal ions.

Zn(s) | Zn2+(aq) || Cu(aq) | Cu(s)

The half cell with the greatest negative potential is on the left of the salt bridge, so Ecell = Eright cell – Eleft cell. In this case, Ecell = +0.34 – -0.76 = +1.10 V.

Eө = -0.76 V Eө = +0.34 V

The left cell is being oxidized while the right is being reduced.


The electrochemical series

The electrochemical series is a list of standard electrode potentials (Eө). The equilibria are written with the electrons on the left of the arrow, i.e. as a reduction.

+0.80Ag+(aq) / Ag(s)

+0.34Cu2+(aq) / Cu(s)

02H+(aq) / H2(g)

-0.76Zn2+(aq) / Zn(s)

-2.36Mg2+(aq) / Mg(s)

Eө / VHalf equationHalf cell

Mg2+(aq) + 2e- Mg(s)

Zn2+(aq) + 2e- Zn(s)

2H+(aq) + 2e- H2(g)

Cu2+(aq) + 2e- Cu(s)

Ag+(aq) + e- Ag(s)

Electrodes with negative values of Eө are better at releasing electrons (i.e. better reducing agents) than hydrogen.


Calculating Ecell

The e.m.f of an electrochemical cell, Ecell, is the difference between the standard electrode potentials of the two half cells.

Ecell = Eө (positive electrode) – Eө (negative electrode)

This can be worked out from the electrode potentials values in the electrochemical series.

The positive electrode is taken to be the least negative half cell, and the negative electrode is the most negative half cell.


Calculating Ecell: worked example

An electrochemical cell is set up using the two half reactions below. What potential difference Ecell would this cell generate?

Ecell = Eө (positive electrode) – Eө (negative electrode)

The zinc half cell has the more negative potential and so forms the negative electrode. Therefore:

Ecell = (+0.34) – (-0.76)

Zn2+(aq) + 2e- Zn(s) Eө = -0.76 V

Cu2+(aq) + 2e- Cu(s) Eө = +0.34 V

= +1.10 V


Calculating Ecell: combining half equations

To find the overall reaction occurring in the cell as a whole, the two half equations are added together:

Because the zinc half cell forms the negative electrode of the cell, oxidation occurs at this electrode and the half equation must be reversed:

The two half equations are added to give the overallcell reaction:

Zn(s) + Cu2+(aq) Zn2+

(aq) + Cu(s)

Zn(s) Zn2+(aq) + 2e-

Cu2+(aq) + 2e- Cu(s)


Calculating Ecell



Predicting the direction of redox reactions (1)

It is possible to use standard electrode potentials to decide on the feasibility of a reaction. Electrodes with more negative electrode potentials have a lower tendency to accept electrons.

Zn(s) + Cu2+(aq) → Zn2+

(aq) + Cu(s)

Zn2+(aq) + 2e- Zn(s) Eө = -0.76 V

Cu2+(aq) + 2e- Cu(s) Eө = +0.34 V

When a pair of electrodes are connected, electrons flow from the more negative to the more positive. The signs of the electrodes can be used to predict the direction of the reaction.

Zn2+(aq) + Cu(s) → Zn(s) + Cu2+

(aq)

feasible

not feasible



Worked example

Fe3+(aq) + e- → Fe2+

(aq)

Electrons flow to the more positive terminal of a cell, which is Fe3+

(aq)/Fe2+(aq). This half cell will accept electrons, a

reduction reaction occurs, and the half equation is:

Step 1: write the equations for the two half reactions:

What reaction would occur if Fe3+(aq)/Fe2+

(aq) and Cu2+(aq)/Cu(s)

half cells were connected?

Fe3+(aq) + e- Fe2+

(aq) Eθ = +0.77 V

Cu2+(aq) + 2e- Cu(s) Eθ = +0.34 V



An oxidation reaction occurs in the other half cell. Electrons are produced and the half equation is:

2Fe3+(aq) + Cu(s) → Cu2+

(aq) + 2Fe2+(aq)

Step 2: combine the two half equations to give a full equation for the reaction.

Cu(s) → Cu2+(aq) + 2e-

The number of electrons supplied and donated must be equal, so the reaction in the Fe3+

(aq)/Fe2+(aq) half cell must

occur twice for each Cu2+(aq)/Cu(s) half cell reaction:

The reverse reaction is not feasible.


Predicting the direction of redox reactions



Glossary


What’s the keyword?


Multiple-choice quiz

1 of 35© Boardworks Ltd 2010. 2 of 35© Boardworks Ltd 2010.

Documents

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oxidized slide

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summary slide

oxidation numbers

balancing equations

assigning oxidation

equations example chlorine