Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 6- C:

The method of half-reaction in basic aqueous solution

Silver is sometimes found as large lumps mixed with other metals in ore form. An easy method to extract

the silver is with the use of the cyanide ion CN- through the following reaction in basic solution:

Ag(s) + 2 CN-(aq) + O 2 (g)  Ag(CN) 2 - (aq)

We will balance the reaction equation using the half-reaction principle. The first step is to identify and

write the reaction equations for the half-reactions. First we write the half-reactions for the oxidation

reaction:

Ag(s) + 2 CN-(aq)  Ag(CN) 2 - (aq) (oxidation)

Similarly for the reduction reaction as follows as we do not know the product from this reaction:

O 2 (g)  ?? (reduction)

The nest step is to balance each of the half-reactions in order to make sure that the charges fit. The

balancing is once again done with electrons. For the oxidation reaction we have:

Ag(s) + 2 CN-(aq)  Ag(CN) 2 - (aq) + e- (oxidation)

We don not know the product of the reaction but from the general rules stated earlier we know that oxygen

is often in the level of oxidation of -2. We thereby assume the product of the reduction reaction is O2-. The

balancing with electron thereby becomes:

4 e- + O 2 (g)  2 O2- (reduction)

As the oxidation involves 1 electron and the reduction involves 4 electrons we multiply the oxidation with

4 and hereby the half-reactions are added giving:

4Ag + 8CN- + O 2  4Ag(CN) 2 - + 2 O2-

In basic solution we have oxide but not on the O2- form but rather on the protonised OH- form giving:

4Ag + 8CN- + O 2  4Ag(CN) 2 - + 2 OH-

The charges of both sides are calculated and balanced:

4×(0) + 8× (-1) + (0)  4×(-1) + 2×(-1) =

-8  -6

Electrochemistry