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.
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^+
6 answers
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^-
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.
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.
Thanks a lot i figured this one out too its: 2H2O + MnO2 + 3Cu^2+ ----> MnO4^- + 3Cu^+ + 4H^+
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.
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3 answers