Balance the following equation for a half reaction that occurs in acidic solution. Use e– as the symbol for an electron.
Mo^3+ -> MoO2^2+
Is the answer 2O+2Mo^3+ -> MOO2^2++3e^-?
No. O should be written O2. Mo doesn't balance. Charge on left and right don't balance. The following are the steps. I think I gave you a site last night to show you how to do this. There are easier ways to do this but I prefer the following steps.
Mo^3+ ==> MoO2^2+
1. Determine the oxidation state of the elements and add electrons to the appropriate side to balance the change Mo is changing; on the left is +3. On the right is +6, so we add 3e to the right.
Mo^3+ ==> MoO2^2+ + 3e
2. Count up the charges on the two sides and add H+ (in acid solution or OH in a basic solution) to balance the charge.
+3 on the left; -1 on the right. Add H^+ to the right.
Mo^3+ ==> MoO2^2+ + 3e + 4H^+
3. Ad H2O to the other side to balance H.
Mo^3+ + 2H2O ==> MoO2^2+ + 3e + 4H^+
4. Check everything.
a. atoms:
1 Mo left and right
2 O left and right.
4 H left and right
b. charge:
+3 on the left; +2 +4 -3 = 3+ on right.
c. change in oxidation state:
Mo^3+ ==> Mo^6+ + 3e
Well, let me tell you, balancing equations is like balancing on a unicycle while juggling balls. It can get pretty tricky! But fear not, because Clown Bot is here to help you with a splash of humor!
To balance the equation, we need to make sure that the number of atoms is the same on both sides of the reaction. Let's give it a shot:
First, let's balance the atoms on the right side of the equation. We have 2 oxygen (O) atoms in the MoO2^2+ ion, so we'll need 2 oxygen (O) atoms on the left side too.
Next, let's balance the charge. In order to balance out the charges, we'll need 3 electrons (e^-) on the left side to counteract the positive charge of the Mo^3+ ion.
So, our balanced equation is: 3e^- + 2Mo^3+ -> MoO2^2+
There you have it! We've balanced the equation and added a little circus flair to make it more fun. Keep practicing those balancing skills, and remember, laughter is the best way to tackle chemistry!
To balance the half reaction:
1. Identify the species undergoing oxidation (loses electrons) and reduction (gains electrons).
In this case, Mo^3+ is being reduced to MoO2^2+, so it is the species undergoing reduction.
2. Write the half reaction for reduction.
Mo^3+ -> MoO2^2+
3. Balance the atoms other than hydrogen and oxygen.
Since there is only one Mo atom on each side of the equation, the Mo atoms are balanced.
4. Balance the oxygen atoms using water (H2O).
Since there are 2 O atoms on the right side, add 2 H2O to the left side:
Mo^3+ + 2H2O -> MoO2^2+
5. Balance the hydrogen atoms using hydrogen ions (H+).
Since there are 4 H atoms on the right side, add 4 H+ to the left side:
Mo^3+ + 2H2O + 4H+ -> MoO2^2+
6. Balance the charges using electrons (e^-).
Since the total charge on the left side is +3 and the total charge on the right side is +2, add 1 electron to the right side to balance the charges:
Mo^3+ + 2H2O + 4H+ -> MoO2^2+ + 1e^-
Therefore, the balanced half reaction in acidic solution is:
Mo^3+ + 2H2O + 4H+ -> MoO2^2+ + e^-
To balance a half-reaction in acidic solution, you need to make sure that the number of atoms and charges on both sides of the reaction are equal. Here's how to balance the given half-reaction:
1. Start by writing the unbalanced half-reaction:
Mo^3+ -> MoO2^2+
2. Balance the atoms:
There is only one Mo atom on each side, so the atoms are already balanced.
3. Balance the charges:
The Mo^3+ ion has a charge of 3+, while the MoO2^2+ ion has a charge of 2+. To balance the charges, you need to add 1 electron (e^-) to the left-hand side of the equation.
Mo^3+ + e^- -> MoO2^2+
4. Multiply the half-reaction by appropriate coefficients to equalize the electrons:
Since you have now added 1 electron to the left-hand side, you need to balance it by multiplying the entire half-reaction by 3.
3(Mo^3+ + e^- -> MoO2^2+)
This gives you the final balanced half-reaction:
3Mo^3+ + 3e^- -> 3MoO2^2+
So, the correct balanced half-reaction for Mo^3+ -> MoO2^2+ in acidic solution would be:
3Mo^3+ + 3e^- -> 3MoO2^2+