5.2.3 (a,b,c) Redox

(a) explanation and use of the terms oxidising agent and reducing agent

(b) construction of redox equations using half-equations and oxidation numbers

(c) interpretation and prediction of reactions involving electron transfer

Half Equations.

Simple Half-equations simply balance charges using electrons.

So, if Iron (II) is Oxidised to Iron (III) we could write:

Fe²⁺ --> Fe³⁺

But the charges on both sides are different.

Adding an electron can put this right

Fe²⁺ --> Fe³⁺ + e^-

The Fe²⁺ was oxidised , it is the reducing agent

But the Iron (II) must have given the electron to some other substance, and this substance must have been reduced because it was forced to gain an electron.

So, if Copper (II) was reduced to Copper (I) in the reaction, the other half equation would be:

Cu²⁺ + e^- --> Cu⁺

The Cu²⁺ was reduced , it is the Oxidising agent

And we can add the two half-equations together and cancel like-terms to make the whole equation:

Fe²⁺ + Cu²⁺ + e^- --> Fe³⁺ + e^- + Cu⁺

Fe²⁺ + Cu²⁺ --> Fe³⁺ + Cu⁺

^{Reducing Iron (V) Ferrate to Iron (II) using Caesium.}

FeO₃^- --> Fe²⁺

We can't balance this without adding extra reactants/products.

The rule is that we can only add H⁺ ions and/or water.

We need 3 O atoms on the right so we add 3 waters

FeO₃^- --> Fe²⁺ + 3 H₂O

But this means we have 6 H atoms on the right, so we 6 H⁺ ions on the left to equalise.

FeO₃^- + 6H⁺ --> Fe²⁺ + 3 H₂O

Finally we have the atoms balanced so we can add electrons.

Total charge Reactants = 5+ Total charge Products = 2+ Add 3 e- on left to balance

3e^- + FeO₃^- + 6H⁺ --> Fe²⁺ + 3 H₂O

Since the Ferrate was reduced the Caesium must be oxidised.

Cs --> Cs⁺ + 1 e^-

Adding the two half-equations together wouldn't cancel out the electrons so we must multiply until they do:

3Cs --> 3Cs⁺ + 3 e^-

Now we can add them together and cancel.

Cs + 3e^- + FeO₃^- + 6H⁺ --> Fe²⁺ + 3H₂O + 3Cs⁺ + 3 e^-

Cs + FeO₃^- + 6H⁺ --> Fe²⁺ + 3H₂O + 3Cs⁺

Dichromate as an Oxidising Agent.

You should know that Dichromate ions are orange and have the formula Cr₂O₇^2-

You should also know that they turn to green Cr³⁺ ions

This allows you to work out the following half equation

Step 1 Cr₂O₇^2- --> 2Cr³⁺

Step 2 Cr₂O₇^2- --> 2Cr³⁺ + 7H₂O

Step 3 Cr₂O₇^2- +14H⁺ --> 2Cr³⁺ + 7H₂O

Step 2 Cr₂O₇^2- +14H⁺ + 6e^- --> 2Cr³⁺ + 7H₂O

Or you could just learn the above.

Clearly, Dichromate would be an Oxidising Agent because it takes electrons away from whatever it reacts with.

When suggesting its use as an Oxidising Agent you must specify Acidified Dichromate because without the H⁺ ions the half-equation will not work.

Manganate (VII) as an Oxidising Agent.

You should know that Manganate (VII) ions are purple and have the formula MnO₄^-

You should also know that they turn to colourless Mn²⁺ ions (in acidic conditions)

This allows you to work out the following half equation

Step 1 MnO₄^- --> Mn²⁺

Step 2 MnO₄^- --> Mn²⁺ + 4H₂O

Step 3 MnO₄^- +8H⁺ --> Mn²⁺ + 4H₂O

Step 2 MnO₄^- +8H⁺ + 5e^- --> Mn²⁺ + 4H₂O

Or you could just learn the above too - although it is less commonly asked in exams than the Dichromate equation.

Clearly, Manganate would be an Oxidising Agent because it takes electrons away from whatever it reacts with.

When suggesting its use as an Oxidising Agent you must specify Acidified Manganate because without the H⁺ ions the half-equation will not work.

Before putting any two half-equations together you must first ensure that the electrons will cancel out.

Eg.

These two equations can't be added together until we equalise the number of electrons.

CH₃CH₂OH + H₂O --> CH₃COOH + 4H⁺ + 4e^- x6 ====> 6CH₃CH₂OH + 6 H₂O --> 6CH₃COOH + 24H⁺ + 24e^-

Cr₂O₇^2- +14H⁺ + 6e^- --> 2Cr³⁺ + 7H₂O x4 ==========> 4Cr₂O₇^2- +56H⁺ + 24e^- --> 8Cr³⁺ + 28H₂O

Now we can add the two left hand sides together and do the same for the right hand sides, cancelling the electrons out.

6 CH₃CH₂OH + 6 H₂O + 4 Cr₂O₇^2- +56 H⁺ + 24 e^- --> 6 CH₃COOH + 24 H⁺ + 24e^- + 8 Cr³⁺ + 28 H₂O

In this case there is water on both sides of the equation, this must be cancelled too.

6 CH₃CH₂OH + 4 Cr₂O₇^2- +56 H⁺ --> 6 CH₃COOH + 24 H⁺ + 8 Cr³⁺ + 22 H₂O

There are Hydrogen ions on both sides of the equation too. Cancel!

6 CH₃CH₂OH + 4 Cr₂O₇^2- +32 H⁺ --> 6 CH₃COOH + 8 Cr³⁺ + 22 H₂O

And all the coefficients are multiples of 2. Cancel!

3 CH₃CH₂OH + 2 Cr₂O₇^2- +16 H⁺ --> 3 CH₃COOH + 4 Cr³⁺ + 11 H₂O