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Jan 19

Balancing redox reactions in basic solutions

By Tutor GuyNo Comments

 

When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how to balance a redox reaction in a basic solution.

Example : Iodide ions are oxidized in a basic solution by permanganate ions to produce iodine and Mn2+ ions. Show the balanced reaction.

Solution: I’ve made this example a little harder than in my other redox reaction posts because we first need to figure out from the example what the half reactions look like. This is not as complicated as it sounds. Just follow the English and turn it into chemical formulas. Our problem should therefore look like this:

I– + MnO4– → I2 + Mn2+

Now comes the tricky part. It turns out it isn’t very easy to balance a redox reaction in basic solutions by adding OH– ions, because when we balance the hydrogen atoms, we throw off the balance of the oxygen atoms. So we employ a clever technique. We first balance the reaction as though it were an acidic solution, then we change the solution from acidic to basic. This is a five-step process, and you’ll notice the first four steps are identical to the process for balancing a redox reaction in an acidic solution:

  1. Determine the number of electrons transferred in each half reaction, and balance the electrons.
  2. Balance the oxygen atoms by adding water molecules.
  3. Balance the hydrogen atoms by adding H+ ions.
  4. Check you answer by verifying that the charges are balanced.
  5. Cancel out the H+ ions by adding an equal number of OH– ions to both sides, which changes the H+ ions into water.

Let’s see how this works on our example. The iodine atom goes from an oxidation number of 1- on the left to 0 on the right, so each iodine atom loses one electron. Since there are two iodine atoms in I2, we think of this half-reaction as losing two electrons. The manganese goes from an oxidation number of 7+ in the permanganate ion to an oxidation number of 2+. It has gained five electrons. To balance the electrons, we need a total transfer of ten electrons. Multiply the iodine compounds by five and the manganese compounds by two. (Note that the iodide ion was multiplied by 10 rather than 5 to balance the number of iodine atoms):

10 I– + 2 MnO4– → 5 I2 + 2 Mn2+

Now we balance the oxygen atoms. There are 8 oxygen atoms on the left and none on the right, so we add 8 water molecules to the right side:

10 I– + 2 MnO4– → 5 I2 + 2 Mn2+ + 8 H2O

Next we balance the hydrogen atoms by adding 16 H+ ions to the left (we pretend we have an acidic solution):

10 I– + 2 MnO4– + 16 H+ → 5 I2 + 2 Mn2+ + 8 H2O

The equation should be fully balanced, and we check our work by verifying the charges balance. 10- (the iodide ions), 2- (the permanganate ions) and 16+ (the hydrogen ions) gives a net charge on the left of 4+. On the right side, we also have a total charge of 4+.

Finally, we are ready to change the reaction from acidic to basic. We cancel out the 16 H+ ions by adding 16 OH– ions to both sides:

10 I– + 2 MnO4– + 16 H+ + 16 OH– → 5 I2 + 2 Mn2+ + 8 H2O + 16 OH–

Because H+ and OH– ions react to form water molecules, we simplify the left side:

10 I– + 2 MnO4– + 16 H2O → 5 I2 + 2 Mn2+ + 8 H2O + 16 OH–

Our last step is to simplify the equation by subtracting the 8 extra water molecules from the right side. We have to subtract the same number from the left side to keep the reaction balanced:

10 I– + 2 MnO4– + 8 H2O → 5 I2 + 2 Mn2+ + 16 OH–

(Look for other tips on this website if you need help with balancing simple redox reactions or redox reactions in acidic solutions.)

Jan 16

Balancing redox reactions in acidic solutions

By Tutor GuyNo Comments

 

When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how to balance a redox reaction in an acidic solution.

Example 1: Balance the following reaction:

MnO4– + NO + H+ → Mn2+ + H2O + NO3–

Solution: To balance reactions like this, use the following four-step procedure:

  1. Determine the number of electrons transferred in each half reaction, and balance the electrons.
  2. Balance the oxygen atoms by adding water molecules.
  3. Balance the hydrogen atoms by adding H+ ions.
  4. Check you answer by verifying that the charges are balanced.

In this reaction, the Mn atom goes from an oxidation number of 7+ in the permanganate ion to an oxidation number of 2+. It has gained five electrons. The nitrogen atom goes from an oxidation number of 2+ in the nitric oxide to 5+ in the nitrate ion. It has lost three electrons. To balance the electrons, we need to make the total number of electrons transferred equal 15 (because that’s the least common multiple of three and five!). We multiply the Mn ions by 3 and the N compounds by 5:

3 MnO4– + 5 NO + H+ → 3 Mn2+ + H2O + 5 NO3–

The next step is to balance the oxygen atoms. There are 17 oxygen atoms on the left side of the reaction and 15 on the right side (before you include the water). Therefore we need two water molecules on the right:

3 MnO4– + 5 NO + H+ → 3 Mn2+ + 2 H2O + 5 NO3–

Finally, we balance the hydrogen atoms (there are four of them in the water molecule on the right) by adding 4 hydrogen ions on the left:

3 MnO4– + 5 NO + 4 H+ → 3 Mn2+ + 2 H2O + 5 NO3–

We check our work by verifying the charges balance. 3- (the permanganate ions) and 4+ (the hydrogen ions) gives a net charge on the left of 1+. 6+ (the manganese ions) and 5- (the nitrate ions) on the right also gives a net charge of 1+. We have succeeded!

Example 2: Complete and balance the following reaction. Assume it takes place in an acidic solution:

Cr2O72– + HSO3– → SO42– + Cr3+

Solution: In this reaction, the Cr atoms go from an oxidation number of 6+ in the dichromate ion to an oxidation number of 3+ in the chromate ion. Therefore, each chromium atom gains three electrons. But since there are two chromium atoms in the dichromate ion, we must think of this half-reaction as gaining six electrons overall. The sulfur atom goes from an oxidation number of 4+ to 6+, so it loses two electrons.

In order to balance the electrons, we need a net transfer of six electrons in each half-reaction. We multiply the ions with the sulfur atoms by three. Note that we also multiply the chromium ion on the right by two to balance the number of Cr atoms:

Cr2O72– + 3 HSO3– → 3 SO42– + 2 Cr3+

The next step is to balance the oxygen atoms.  There are 16 oxygen atoms on the left and 12 on the right, so we add 4 water molecules to the right side:

Cr2O72– + 3 HSO3– → 3 SO42– + 2 Cr3+ + 4 H2O

Finally, we balance the hydrogen atoms by adding H+ ions. There are three H atoms on the left and 8 H atoms on the right, so we need five H+ ions on the left:

5 H+ + Cr2O72– + 3 HSO3– → 3 SO42– + 2 Cr3+ + 4 H2O

We check our answer by checking the charge on each side. The left side has charges of 5+ (the H ions), 2- (the dichromate ion) and 3- (the hydrogen sulfite ions) for a net of zero. The right side has charges of 6- (the sulfate ions) and 6+ (the chromate ions) also for a net of zero.

(Look for other tips on this website if you need help with balancing simple redox reactions or redox reactions in basic solutions.)

Jan 12

Balancing redox reactions– the basics

By Tutor GuyNo Comments

 

When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how the process works when both the oxidizing agent and the reducing agent are single elements.

Example (a very simple example): Balance the following reaction:

… Mg(s) + … Al3+(aq) → … Mg2+(aq) + … Al(s)

Solution: At first glance, it may appear that the reaction is already balanced because there is one magnesium atom and one aluminum atom on each side. But notice that the charges do not balance. Because aluminum gains three electrons in this reaction, but magnesium atoms lose only two electrons, you will need more magnesiums than aluminums. To find the right ratio, determine the least common multiple of 2 and 3, which is 6. Then multiply the magnesiums by 3 and the aluminums by 2 so that there is a total transfer of 6 electrons in this reaction:

3 Mg(s) + 2 Al3+(aq) → 3 Mg2+(aq) + 2 Al(s)

The charges balance, as required.

(Look for other tips on this website if you need help with balancing redox reactions in acidic or basic solutions.)

Jan 09

Balancing simple chemical reactions

By Tutor GuyNo Comments

 

When balancing equations, start with the atoms that show up in only one compound on each side. Figure out those coefficients first, and then work back and forth from the reactants side to the products side to find the coefficients of each of the other compounds.

Example 1: Balance the following equation:

 … C5H12 + … O2 → … CO2 + … H2O

 Solution: This reaction showing the combustion of pentane is not too complicated. Notice that carbon appears in only one compound on each side, so it’s a good place to start. The five carbon atoms on the left need to be balanced by five carbon atoms on the right:

 1 C5H12 + … O2 → 5 CO2 + … H2O

 There are 12 hydrogen atoms on the left, so balance these on the right:

 1 C5H12 + … O2 → 5 CO2 + 6 H2O

 Finally, count up all the oxygen atoms on the right (there are 16 of them!) and balance with the O2 on the left:

 1 C5H12 + 8 O2 → 5 CO2 + 6 H2O

 Example 2: Balance the following equation:

 … C6H14 + … O2 → … CO2 + … H2O

 Solution: Changing the pentane to hexane has made this equation more complicated. Let’s see why. First, let’s balance the carbon and hydrogen as before:

 1 C6H14 + … O2 → 6 CO2 + 7 H2O

 There are now 19 oxygen atoms on the right side. How do you balance this on the left side? Some teachers will let you use fractions for your coefficients, and 19/2 is the correct fraction in this case. But most teachers want you to use only whole numbers for the coefficients. If you encounter a fraction, you need to multiply the entire equation by the denominator to finish the problem. Here, we multiply the equation above by 2:

 2 C6H14 + … O2 → 12 CO2 + 14 H2O

 Notice there are now 38 oxygen atoms on the right side, and the last step is simple:

 2 C6H14 + 19 O2 → 12 CO2 + 14 H2O

 Here’s a trick you can apply when the equation is a double replacement reaction:

Example 3: Balance the following reaction:

 … Na2SO4 + … Al(NO3)3 → … NaNO3 + … Al2(SO4)3

Solution: Let’s start by balancing the aluminum:

 … Na2SO4 + 2 Al(NO3)3 → … NaNO3 + 1 Al2(SO4)3

 Now instead of balancing the sulfur atoms, we can save time by balancing the sulfate ions instead. We treat them as a group rather than as separate atoms. There are three sulfates on the right side, so we need three on the left:

 3 Na2SO4 + 2 Al(NO3)3 → … NaNO3 + 1 Al2(SO4)3

 Note that you can only balance polyatomic ions like this when the ions do not change from reactant side to product side. Finally, you can see there are six sodium atoms on the left (or, six nitrate ions), so the final answer is:

 3 Na2SO4 + 2 Al(NO3)3 → 6 NaNO3 + 1 Al2(SO4)3

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