In a particular redox reaction, MnO2 is oxidized to MnO4– and Cu2 is reduced to Cu . Complete and balance the equation for this reaction in acidic solution. Phases are optional.
MnO2 + Cu^2+ ---> MnO4^- + Cu^+
This is how I am doing it:
1. break up into half reactions
MnO2--->MnO4 and Cu^2+ ---> Cu^+
2. add H20, H+, and e- to both sides and I get:
2H2O + MnO2 ---> MnO4 + 4H + 8e^- and
e^- + Cu^2+ ----> Cu^+ then adding them I get:
2H2O + MnO2 + Cu^2+ ---> MnO4 + Cu^+ + 4H^+ + 7e^-
I think the trouble you are having is caused by not putting the charges in. That H on the right should be 4H^+. The MnO4 on the right should be MnO4^-. When you do that the e need on the right is not 8 but 3 so the charge on the left is zero and on the right is (-1+4-3) = zero.
Then you multiply both equation to make the change in e equal. Tha tmeans you multiply equation 1 by 1 and equation 2 by 3 then add everything.
This one is right. If you haven't already do so go back and look at the sulfite ==> sulfate because the one is not right.
Chemistry and statistic
And physics
To balance the redox equation in acidic solution, you need to follow these steps:
Step 1: Assign oxidation numbers to each element in the equation.
MnO2: Mn has an oxidation number of +4, and O has an oxidation number of -2.
MnO4-: Mn has an oxidation number of +7, and O has an oxidation number of -2.
Cu^2+: Cu has an oxidation number of +2.
Cu+: Cu has an oxidation number of +1.
Step 2: Separate the equation into two half-reactions:
Half-reaction 1: MnO2 -> MnO4-
Half-reaction 2: Cu^2+ -> Cu
Step 3: Balance the atoms and charges in each half-reaction.
Half-reaction 1:
MnO2 -> MnO4-
To balance the number of Mn atoms, you need to add 2H+ to the reactant side:
MnO2 + 2H+ -> MnO4-
Next, balance the number of oxygen atoms by adding water (H2O) molecules to the reactant side:
MnO2 + 2H+ -> MnO4- + 2H2O
Check the charge on both sides, and you'll find that both sides have a charge of -2, balancing out the equation.
Half-reaction 2:
Cu^2+ -> Cu
The number of Cu atoms is already balanced, so now we need to balance the charges. Since Cu^2+ has a charge of +2, and Cu has a charge of +1, we need to add one electron (e-) to the product side:
Cu^2+ + 1e- -> Cu
Step 4: Multiply the half-reactions by appropriate coefficients to equalize the number of electrons transferred.
To do this, we need to multiply the first half-reaction by 1, and the second half-reaction by 2:
2(MnO2 + 2H+ -> MnO4- + 2H2O)
Cu^2+ + 2e- -> Cu
Step 5: Combine the half-reactions.
Add the two half-reactions together:
2(MnO2 + 2H+ -> MnO4- + 2H2O)
Cu^2+ + 2e- -> Cu
--------------------------
2MnO2 + 4H+ + Cu^2+ + 2e- -> 2MnO4- + 4H2O + Cu
Step 6: Simplify the equation by canceling out any common terms.
Final balanced equation in acidic solution:
2MnO2 + 4H+ + Cu^2+ + 2e- -> 2MnO4- + 4H2O + Cu
Thanks a lot i figured this one out too its: 2H2O + MnO2 + 3Cu^2+ ----> MnO4^- + 3Cu^+ + 4H^+
Again you need to learn to do these. Mn on the left is +4 and on the right is +7. Cu on the left is 2+ and on the right is zero. Balance the electron change first. Mn on the left goes to Mn on the right with ? change of e.
Cu on the left goes to Cu on the right with ? change in e.