The following redox equation occurs in acidic aqueous solution. When it is balanced with the smallest whole number coefficients, what is the coefficient for chlorine gas?

KMnO4(aq) + KCl(aq) + H2SO4(aq) = MnSO4(aq) + K2SO4(aq) + Cl2(g)

a. 1
b. 3
c. 5
d. 8
e. 10

I'm not going to balance this thing and give you the answer. This will get you started.

Mn goes from oxidation state of +7 on the left to +2 on the right.
Cl goes from -1 on the left to zero on the right (for each Cl atom/ion).
Hint: Temporarily balance the Cl atoms by placing a 2 in front of KCl and figure the electron change for two Cl^- going to Cl2. You may or may not need to change this two later.
If you are lost with this, tell us what your problem is and we can help you through it. You need to learn how to do these yourself.

Hi doctor,

I have a Chemistry test coming up this Friday with 30 Multiple Choice. It's going to be mostly calculative problems from 11 chapters. I'm just not quite sure how my approach to studying should be. Can you give me any tips? Much appreciated, thanks.

Here is a very good site that takes you through determining the oxidation state of elements (alone and in compounds) as well as how to write the half reactions and how to balance the complete equation.

(Broken Link Removed)

To balance this redox equation in acidic solution, we need to follow a few steps:

Step 1: Identify the elements that are undergoing changes in oxidation state. In this equation, we can see that Mn is reducing and Cl is oxidizing.

Step 2: Write down separate half-reactions for each element involved. In this case, we have:

Half-reaction for Mn:
MnO4^- → Mn^2+

Half-reaction for Cl:
Cl^- → Cl2

Step 3: Balance the atoms in each half-reaction, excluding oxygen and hydrogen. For the Mn half-reaction, we have:

MnO4^- → Mn^2+
Balancing Mn atoms:
MnO4^- → Mn^2+

Step 4: Balance the oxygen atoms in each half-reaction by adding water (H2O) molecules. In this case, we need to add four water molecules to balance the oxygen atoms in the Mn half-reaction:

MnO4^- + 4H2O → Mn^2+

Step 5: Balance the hydrogen atoms in each half-reaction by adding hydrogen ions (H+). In this case, we need to add eight hydrogen ions to balance the hydrogen atoms in the Mn half-reaction:

MnO4^- + 8H+ + 4H2O → Mn^2+

Step 6: Balance the charges in each half-reaction by adding electrons (e^-). In this case, we need to add five electrons to the left-hand side to balance the charge:

MnO4^- + 8H+ + 4H2O + 5e^- → Mn^2+

Now let's balance the Cl half-reaction:

Cl^- → Cl2

Step 7: Balance the atoms in the Cl half-reaction. In this case, we have:

2Cl^- → Cl2

Step 8: Combine the two half-reactions by multiplying so that the electrons cancel out. In this case, we need to multiply the Mn half-reaction by 2 and the Cl half-reaction by 5:

2MnO4^- + 16H+ + 8H2O + 10e^- → 2Mn^2+ + 10e^-
10Cl^- → 5Cl2

Step 9: Add the two half-reactions together and simplify if necessary:

2MnO4^- + 16H+ + 8H2O + 10e^- + 10Cl^- → 2Mn^2+ + 10e^- + 5Cl2

Cancel out the electrons:

2MnO4^- + 16H+ + 8H2O + 10Cl^- → 2Mn^2+ + 5Cl2

Finally, we get the balanced equation:

2KMnO4 + 16KCl + 8H2SO4 → 2MnSO4 + 8K2SO4 + 5Cl2

Now we can see that the balanced coefficient for chlorine gas (Cl2) is 5.

Therefore, the correct answer is option c: 5.