writing half reactions of a single displacement reaction

Question

2AgNO3 (aq) + Cu (s) --> 2 Ag (s) + Cu(NO3)2 (aq)

Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction. 

Answer 1

In a single replacement reaction, the oxidation and reduction occur between the lone molecules.

When something is in elemental form (solid metal, diatomic gas) it’s oxidation number is 0.

Copper begins in its elemental form as solid copper, then ends as part of the aqueous solution. Silver begins as part of the aqueous solution and ends as a solid in elemental form. These two metals switching between solid and aqueous are the reason this is a redox reaction.

Let’s determine the charge of Ag and Cu on the reactant side versus product side. This requires knowledge of the periodic table and charges of ions. In general, for groups 13-17 the charge is

Group # - 18 for negative charges

or

One’s digit of Group # for positive charges

For transition metals such as Ag and Cu, we have to determine the charge of EVERYTHING ELSE and then determine what their charges should be so the compound comes out to have the correct charge.

Reactants

AgNO3 = neutral charge of 0 for whole compound

O = -2 charge (16-18 = -2) and we have 3 oxygens, so in total they contribute -6 charge to the compound

N = -3 or + 5 (15-18 = -3 or group 15 = +5)

N should be positive to help fight the -6 Oxygen

Ag = ?

O gives -6 and N gives +5 so to reach 0 charge the Ag must contribute +1

Ag = +1

Cu = elemental form = 0

Products:

Cu(NO3)2 = neutral charge of 0 for whole compound

O = -2 charge and there are 6 oxygens (3 * 2 = 6) so we have a total charge of -12 from the O’s

N = + 5 and there are 2 N’s (1*2=2) so we have a total charge of +10 from the N’s

If O is -12 and N is +10, the Cu needs to be +2 to help get the compound to 0 neutral charge

Cu = +2

Ag = 0 in elemental form

The change that occurred from reactants to products is

Ag = +1

Cu = 0

Ag = 0

Cu = +2

Oxidized = whoever ended up more positive than when they started = Cu went from 0 to 2

Reduced = whoever ended up less positive than when they started = Ag went from 1 to 0

Oxidation reaction is the reaction showing that this atom gave away electrons so it could become more positive.

Copper gave away 2 electrons to become Cu^2+

Cu(s) —> Cu^2+ + 2e-

Reduction reaction shows the atom that accepted the electrons to become more negative:

Silver accepted 1 electron to become Ag(s)

Ag^+ + 1e- —> Ag(s)

Lastly, we have to make sure this exchange is equal and all electrons are accounted for. If copper gave away 2, silver must also accept 2. We need to take the whole silver equation and multiply everything by 2 so it remains balanced while also accounting for all of the electrons being given away:

2Ag^+ + 2e- —> 2Ag(s)

The equations are:

Cu(s) —> Cu2+ + 2e- oxidation

2Ag+ + 2e- —> 2Ag(s) reduction