Chemistry Lesson: Part 7 (Aqueous Chemical Reactions 2)
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Introduction
This series of posts seeks to present the material covered in the first semester of a college level general chemistry course, in an easily digestible steemit blog post format. The series is intended to be read, and experienced in sequential order starting with Post 1. The material will build upon itself, and potential exercises included (problem sets), will pertain to the post they are contained in, or any previous post. Each post will pick up immediately where the previous in the series left off. Please check out the #chemistry-lesson tag for all posts in this series. I hope you find this series to be informative and beneficial toward your understanding of chemistry and science in general.
Immediate Preceding Post
Part 6: Aqueous Chemical Reactions 1
Legend for This Section
As subscripts do not currently work on steemit, to symbolize when a number should be sub-scripted I will be writing it in the text as follows: /x/.
Precipitation Reactions
Having covered Acid/Base reactions in the previous section, it’s time we move on to another type of reaction, the precipitation reaction. Precipitation (no not the weather kind) is just the formation of an insoluble product. When we talked about a salt getting hydrated by water and splitting up into its ions that happens because the compound is soluble (able to get pulled into the aqueous phase by the solvent, H/2/O). A precipitation reaction is a chemical reaction which forms an insoluble product that crashes out of the solution. When that happens the solution will turn cloudy:
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The above reaction is one between Nickel and Sodium Hydroxide, this reaction forms the insoluble precipitate Nickel (II) Hydroxide or Ni(OH)/2/ the precipitation of which is what gives the solution in the test tube above, its cloudy look.
Redox Reactions
Redox or Oxidation/Reduction reactions are those where electrons get transferred as the components interact. To understand these types of reactions we must think about an element versus its ion. Lets take sodium metal (Na) and its ion Na+ as an example to start:
For sodium metal to form into its ion, it must release an electron, as described in the reaction above. In an oxidation-reduction reaction the electrons that are produced from forming ions of one compound are transferred to the other component. Let’s take the reaction between a metal and an acid as an example:
Here we have the reaction between Magnesium and Hydrochloric Acid. This reaction produces Magnesium Chloride and Hydrogen gas. When dealing with redox reactions we must first think about the ions as I described above. Magnesium splits into Mg2+ and two electrons when it forms its ion. This reaction is called a half reaction because it is the one half of the redox reacton (the part producing the electron). Next we know that aqueous Hydrochloric Acid splits into a Cl- ion and an H+ ion. Since an electron is being transferred and we are producing H/2/ gas, the H+ Ions are the ones accepting those electrons. This is the other half reaction as it is the second half of the redox reaction (the part accepting the electron). So now we have all of the necessary pieces to put this reaction back together. When we do this we see that we have electrons on either side and a total ionic equation (because it lists all of the ions for both the products and reactants). We can see that we have the same number of electrons on both sides of the reaction so we can cancel them out. Now we can also recombine the ionic species to tidy things up, and behold we have re assembled the reaction after having detailed out where the electrons were moving from and to where they were going.
Oxidation Numbers
So how do I know how many electrons are produced when an element becomes an ion? I know this because there is a system that I can use to help me determine the oxidation number of the pieces in a compound. An oxidation number refers to the charge an atom would have were it to form into an ion. Here are some rules for assigning oxidation numbers:
- Oxygen (almost) always has an oxidation number of -2
- Hydrogen (almost) always has an oxidation number of +1 (except when bound to a metal then its -1 eg NaH called sodium hydride)
- Group 1 Metals (Li, Na, K, Rb, Cs) have an oxidation number of +1
- Group 2 Metals (Be, Mg, Ca, Sr, Ba) have an oxidation number of +2
- Halogens (F, Cl, Br, I, At) have an oxidation number of -1
- All elemental compounds (O/2/, H/2/ etc) have an oxidation number of 0
So let’s assign some oxidation numbers to compounds and use them to determine the overall charge of the ion:
Here we have phosphoric acid, let us use the rules to assign oxidation numbers. Oxygen has an oxidation number of -2 and we have four of them, so the total charge from oxygen is -8. Hydrogen has an oxidation number of +1 when not bound to a metal and there are 3 of them so the total charge from Hydrogen is +3. This leaves us with Phosphate, since the total charge on the compound is neutral, the oxidation number of the phosphate must be such that it balances out the charge, so far we have a charge of (-8 + +3) = -5 from the other elements in the compound. This means the oxidation number of the phosphate must be +5.
Here we have strontium nitrate, which you may have some familiarity with if you have ever used a road flare, this is used to make the flares burn red:
When we see a portion of a compound in parentheses this means that portion is a polyatomic ion. In this case it is the Nitrate ion and has a net charge of -1. Let’s assign the oxidation numbers. Oxygen has an oxidation number of -2 and we have three of them, so the total charge from oxygen is -6. The polyatomic Nitrate ion has a charge of -1, so the oxidation number of Nitrogen must be +5 as it allows for the ion to have its charge (-6 + +5 = -1). We also know that we have two Nitrates, and the total charge on our compound is zero. This means that Strontium must have an oxidation number of +2 to balance out the two negative charges from the Nitrates.
End of Aqueous Chemical Reactions 2 Problem Set
We’ve reached the end of this section, and as with the end for previous sections, I will provide you with a problem set so you could test yourself to see if you are fully understanding the material.
Problem Set 6
Problem Set 6 Answer Key
Future Posts
Subsequent posts will cover: Electronic Configuration of Atoms, Chemical Bonding, and Molecular Geometry, and more.
Reference Figure: Periodic Table
Other References
Constants and Conversions List
Source for Additional Constants
Some Common Ions
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This series is an extremely useful resource of knowledge for someone like me, thank you!
I am glad you are enjoying it, comments like this are what keep me going at it!
+1 to that!
Great lesson. It brings me back to the college days!
Thanks! That means I must be doing something right! :)
I remember now! :)
Maybe could the first picture (the one with the details on the MG + 2HCl reaction) be simplified? It looks a bit confusing to me. I would start with eqs 1, 2 and 3. Then you inject equations 2 and 3 into equation 1, introduce equation 4 and get to the final result. But this is at the end of the day a matter of taste, I guess :)
I will be coming back to that sort of material at some point and going into it with more detail. I will teach more about the half reactions and the specific steps I take to balance redox equations at that point! :) Here I was just trying to take a reaction, spread it into its pieces and put it back together again.
Really enjoyed our post!! thank you so much for sharing!
Thank you for the kindness!
Easy and helpful ! Thank you as always :)
I am glad you enjoyed it!
This is a very good remembering lesson!! Thank you @justtryme90
Thank you for your kind words! This is what keeps me going!
Edits:
Edit 1 - Corrected the equation for ionization of Sodium, was not changed to an image file.
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