Google
 

Solution Reactions


Precipitation Reactions

Precipitate - an insoluble solid formed by a reaction in solution.

Consider the following molecular equation:

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

The net ionic equation of the above molecular equation is:

Ag+(aq) + Cl-(aq) AgCl(s)

In order to know which ions will combine to form precipitates, we have the solubility rules

Solubility Rules

  1. All salts containing NH4+ and group IA cations (Li+, Na+, K+, Rb+, Cs+, ) are soluble.
  2. All salts containing NO3-, C2H3O2-, HClO3-, and ClO4- are soluble.
  3. All salts containing Cl-, Br-, and I- are soluble, except those with Ag+, Hg22+, and Pb2+
  4. All salts containing SO42- are soluble, except PbSO4, BaSO4, HgSO4, CaSO4, and AgSO4.
  5. Most salts containing O2-, OH-, PO43-, CO32-, and S2- are insoluble, except those containing NH4+ and group IA cations.

Determine the net ionic equation for the following reaction:

Ba(NO3)2 + Na2SO4 BaSO4 + 2NaNO3

Rule number 4 tells us that BaSO4 is insoluble. So we can write the complete ionic equation as

Ba2+(aq)+2 NO3-(aq) +2 Na+(aq)+SO42-(aq) BaSO4(s)+2 Na+(aq)+NO3-(aq)

Removing the spectator ions leaves us with the net ionic equation:

Ba2+(aq) + SO42-(aq) BaSO4(s)





Gas Reactions

Sometimes a gas will be involved as one of the reactants or products in a solution reaction. For example,

2 HCl(aq) + Na2S(aq) H2S(gas) + 2 NaCl(aq)

In this example the complete ionic equation would be:

2 H+(aq) + 2Cl-(aq) + 2Na+(aq) + S2-(aq) H2S(g) + 2Na+(aq) + 2Cl-(aq)

Removing the spectator ions we obtain the net ionic equation:

2 H+(aq) + S2-(aq) H2S(g)

H2S is just one example of a gaseous substance that can form in a solution reaction. Another way gases can form in solution is through the decomposition of weak electrolytes. For example, H2CO3 readily decomposes into H2O and CO2 gas,

H2CO3(aq) H2O(l) + CO2(g)

So, any solution reaction that leads to the production of H2CO3, such as

HCl(aq) + NaHCO3(aq) NaCl(aq) + H2CO3(aq)

will ultimately lead to the production of H2O and CO2 gas as the H2CO3 decomposes, giving us a new equation:

HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g)

If we eliminate all the spectator ions we would write the net ionic equation as

H+(aq) + HCO3-(aq) H2O(l) + CO2(g)

Two other substances that will decompose and form gases are H2SO3 and NH4OH:

H2SO3(aq) H2O(l) + SO2(g)

NH4OH(aq) H2O(l) + NH3(g)





Weak Electrolyte and Water Formation

Because weak electrolytes do not completely dissociate into ions, whenever two ions can get together to form a weak electrolyte there will be a net ionic equation. For example,

HCl(aq) + NaC2H3O2(aq) HC2H3O2(aq) + NaCl(aq)

Writing out the complete ionic equation we obtain:

H+(aq) + Cl-(aq) + Na+(aq) + C2H3O2-(aq) HC2H3O2(aq) + Na+(aq) + Cl-(aq)

Removing the spectator ions we obtain the net ionic equation:

H+(aq) + C2H3O2-(aq) HC2H3O2(aq)

Water is also considered to be a weak electrolyte. That is, only a small fraction of the H2O molecules in water dissociate to form H+ and OH- ions. Therefore, any reaction that leads to the formation of water will have a net ionic equation. Let's look at a classic example of an acid reacting with a base.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Removing the spectator ions we obtain the net ionic equation:

H+(aq) + OH-(aq) H2O(l)





Reduction/Oxidation Reactions

Any reaction where electrons are transferred between reactants is called an reduction-oxidation reaction or redox reaction. These are reactions where one substance wants an electron so badly that it takes it away from another substance. Whether or not it succeeds depends on who it meets. For example, consider the reaction

FeCl2(aq) + CeCl4(aq) FeCl3(aq) + CeCl3(aq)

What happened here? If we remove the spectator ions and write the net ionic equation we find:

Fe2+(aq) + Ce4+(aq) Fe3+(aq) + Ce3+(aq)

Ce4+ took an electron from Fe2+! This is an oxidation/reduction reaction. In this example, Fe2+ is oxidized and Ce4+ is reduced. The charge of Fe went from +2 to +3, that is, it lost an electron. This process is called oxidation.

Oxidation: The loss of an electron by a substance.

Likewise, the charge of Ce went from +4 to +3, that is, it gained an electron. This process is called reduction.

Reduction: The gain of an electron by a substance.





Half-Reactions

Oxidation/reduction reactions are important because we can exploit them as a way of generating electrical current. For example, we know that Ce4+ will pull an electron away from Fe2+ when we mix the two in solution. The trick to making a battery is to find a way to make Ce4+ pull an electron from Fe2+ when they are not mixed together in a single solution.

In order to do this we set up two 1/2 reactions in separate beakers and connect them with a salt bridge. The salt bridge electrically connects the two beakers, but prevents Fe2+ and Ce4+ from mixing.

alternatetext

You need two 1/2 reactions to make a reaction, so, we add them together.

Fe2+ Fe3+ + e-
Ce4+ + e- Ce3+
__________________________________
Fe2+ + Ce4+ Fe3+ + Ce3+

Notice that electrons on both sides of the half-reactions must cancel each other out when added together.

What about other atoms and molecules. How do you know if one chemical substance is strong enough to take an electron from another? We simply refer to a list known as the activity series.





Activity Series

The activity series is a list of metals and their half-reactions arranged in order of decreasing ease of oxidation or increasing ability to take an electron.

alternatetext

How do you use this series? Very simply, any substance on the right-hand side of the arrows will be strong enough to take an electron from any substance above it on the left-hand side of the arrows.

For example, if a Ag+ cation gets close to an Fe atom then the Ag+ cation will take an electron away from the Fe atom. In terms of a chemical reaction the activity series predicts

2Ag+ + Fe 2 Ag + Fe2+

In contrast, when a Fe2+ cation meets a Ag atom, the Fe2+ cation will not be strong enough to take electrons away from the Ag atom. That is,

2 Ag + Fe2+ X 2Ag+ + Fe

What happens when we mix Ca metal and HCl acid?

Ca(s) + HCl(aq) ?

First, we write the complete ionic equation for the reactants.

Ca(s) + H+(aq) + Cl-(aq)

Then we ask if H+(aq) is strong enough to take an electron away from Ca(s)?

We look at the activity series and see that H+ is below Ca so H+(aq) can indeed take an electron from Ca(s). Now we can write the two half-reactions for this reaction. Since H+(aq) gains electrons, it undergoes reduction, so we write its half-reaction as

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

And, since Ca loses electrons, it undergoes oxidation, so we we write its half-reaction as

Ca(s) Ca2+(aq) + 2e-

Finally to get a net ionic equation we add these two half-reactions together to obtain:

Ca(s) + 2H+(aq) Ca2+(aq) + H2(g)

Again making certain that all the electrons on both sides of the equation cancel. After putting the spectator ions back in we can write the balanced molecular equation:

Ca(s) + 2HCl(aq) CaCl2(aq) + H2(g)

Using the activity series can you explain why the metals gold, silver, platinum, and palladium are so highly valued and called the precious metals?





Generally, many metals will react with acids to form a salt and hydrogen gas.

Metal + Acid Salt + H2(g)

Let's look at some other examples. What do you expect will happen when Zn metal is put into a beaker of HCl(aq)

Using the activities series we can predict:

Zn(s) + 2 HCl(aq) Zn2+ + 2 Cl- + H2(g)

how about when Cu metal is put into a beaker of HCl(aq)? In this case the activity series tell's us there's no reaction. So in summary, by simply looking at the Activity Series you can predict which redox reactions will occur and which won't.

What will happen when copper wire is placed into a AgNO3 solution?

The complete ionic equation for the reactants is

Cu(s) + Ag+(aq) + NO3-(aq) ?

From the activity series we see that Ag+ is stronger than Cu so a reaction will occur. The half-reactions are:

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

Homework from Chemisty, The Central Science, 10th Ed.

4.19, 4.21, 4.23, 4.25, 4.27, 4.39, 4.41, 4.43, 4.45, 4.47, 4.49, 4.51, 4.53, 4.55, 4.57

Previous Next

                                    |