In a particular redox reaction, BrO– is oxidized to BrO3– and Cu2 is reduced to Cu . Complete and balance the equation for this reaction in acidic solution.
Two half reactions:
1.) Cu^2+ --> Cu^+
2.) BrO^- --> BrO3^-
Balanced:
1.) Cu^2+ +e^- --> Cu^+
2.) BrO^- + 2H2O--> BrO3^- + 4H^+ 3e^-
Cross multiply (so e^- cancel):
1.) 3Cu^2+ + 3e^- --> 3Cu^+
2.) BrO^- + 2H2O--> BrO3^- + 4H^+ 3e^-
Combine & cancel:
BrO^- + 3Cu^2+ + 2H2O --> BrO3^- + 3Cu^+ + 4H^+
Not sure what I'm doing wrong but my answer is not right. Thank you for help!
On #2, atoms balance but charge does not balance. Also, change of electrons is not balanced. Br is +1 on the left and +5 on the right; therefore, change in electrons must be 4 and not 3
That was it! As always, thank you :)
This answer is still not correct for me!!
To balance the redox reaction in acidic solution, we need to make sure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. Let's analyze the given half-reactions:
1.) Cu^2+ → Cu^+
2.) BrO^− → BrO3^−
First, let's balance the reduction half-reaction (Cu^2+ → Cu^+). Since copper is being reduced, it gains one electron:
Cu^2+ + e^− → Cu^+
Next, let's balance the oxidation half-reaction (BrO^− → BrO3^−). Since bromine goes from a lower oxidation state in BrO^− to a higher oxidation state in BrO3^−, it loses three electrons:
BrO^− → BrO3^− + 3e^−
Now, let's multiply each half-reaction by the appropriate factor so that the number of electrons gained in the reduction equals the number of electrons lost in the oxidation:
3(Cu^2+ + e^− → Cu^+)
2(BrO^− → BrO3^− + 3e^−)
This gives us:
3Cu^2+ + 3e^− → 3Cu^+
2BrO^− + 6e^− → 2BrO3^−
Now, we can add the two half-reactions together and cancel out common terms:
3Cu^2+ + 3BrO^− + 2H2O → 3Cu^+ + 2BrO3^− + 6H^+
Therefore, the balanced redox reaction in acidic solution is:
3Cu^2+ + 3BrO^− + 2H2O → 3Cu^+ + 2BrO3^− + 6H^+