Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107	Fundamentals of Chemistry	Fall 2008
Lecture Notes: 23 October	© R. Paselk 2005
PREVIOUS		NEXT

Chemical Reactions

Net Ionic Equations, cont.

Let's look at a couple of more examples of net ionic equations:

• Consider the reaction of 50.0 mL of 0.25 M hydrochloric acid with 25.0 mL of 0.50 M sodium hydroxide.
  • Write a net ionic equation for this reaction:

H⁺_(aq) + Cl^-_(aq) + Na⁺_(aq) + OH^-_(aq) H₂O + Cl^-_(aq) + Na⁺_(aq)

Giving: H⁺ + OH^- H₂O

• How much acid will be left over? Base? How much water was made (synthesized)?
  • First find out the moles of each: acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10^-2moles; base = (25.0 mL)(1 L/1000 mL)(0.50 mole/L) = 1.25 x 10^-2moles.
  • Since the amounts are identical they will completely neutralize each other. That is they will react completely and both acid and base will be consumed with none left over.
  • Since the ratio in the equation is 1:1:1 acid:base:water, have 1.25 x 10^-2moles = 1.3 x 10^-2moles (or 1.2 x 10^-2moles) of water produced.
• Consider the reaction of calcium metal with hydrochloric acid. This is similar to the reaction of sodium and water, and hydrogen gas is given off.
  • Write a net ionic equation for this reaction:

H⁺_(aq) + Cl^-_(aq) + Ca⁰_(s) H_{2 (g)} + Ca²⁺_(aq) + Cl^-_(aq)

Balancing: 2 H⁺ + Ca⁰_(s) H_{2 (g)} + Ca²⁺

• Notice that this net ionic equation also involves electron transfer, so it is a redox equation.

Oxidation/Reduction Reactions

In these reactions we see a transfer of electrons from one atom or molecule to another. First let's look at some terms.

• Oxidation refers to taking electrons away from a substance. So to oxidize means to behave like oxygen normally does and "steal" electrons.
• Reduction refers to receiving electrons - it is the opposite of oxidation.
• Note that oxidation and reduction always go together. When oxygen oxidizes it is itself reduced (it gains electrons).
• Consider the example of burning natural gas:

CH₄ + 2 O₂ CO₂ + 2 H₂O

Notice that the methane is oxidized by the oxygen. We say that the carbon and hydrogen are both oxidized to give the new covalent products, water and carbon dioxide.

Examples:

• A piece of clean copper is placed in a solution of silver nitrate. Assuming the copper goes to Cu(II) write a balanced equation. Notice that in this case you must balance the charge in order to get the final equation.

Ag⁺ + NO₃^- + Cu⁰ Ag⁰ + NO₃^- + Cu²⁺

Balancing: 2Ag⁺ + Cu⁰ 2Ag⁰ + Cu²⁺

• A piece of clean iron is placed in a solution of copper(II) sulfate. Assuming iron goes to Fe(III) write a balanced equation. Again the problem is to balance the charge.

Cu²⁺ + SO₄^2- + Fe⁰ Cu⁰ + Fe³⁺ + SO₄^2-

Balancing: 3 Cu²⁺ + 2 Fe⁰ 3 Cu⁰ + 2 Fe³⁺

As another example we can look at a key oxidation reaction in glycolysis, the central pathway of metabolism. Don't worry about these reactions - they will not be on an exam. They are presented for your interest. In this case we use a biological oxidizing agent, NAD+ to take electrons (in the form of a hydride ion, H:-) Frequently organisms need to operate this reaction under anaerobic (oxygen free) conditions. For example the maximum power (energy/sec) you can get from your muscles is anaerobically. For this to occur you need to regenerate the oxidizer (NAD+) without the presence of oxygen. For this to occur we use another reaction: The mechanism for this electron transfer is shown below:

C107 Laboratory

C107 Home

C107 Lecture Notes

© R A Paselk

Last modified 23 October 2008