I have some challenging homework of balancing redox equations! I get really frustrated from it and I have a quiz 2 days later! Please help me out!
*When dealing with acidic/basic conditions, we write half reactions as a compound so that we can balance O by addiong H2O and balance H by add H+.
Balance the following equations using half-reactions and oxidation numbers:
1)H2O2 + Cr2O7 2- -> Cr 3+ + O2 + H2O (ACIDIC condition)
oxidation numbers are +1,-1,+6,-2,->,+3,0,+1,-2 respectively
Half reactions:(not balanced)
Cr2O7 2- -> 2Cr 3+
H2O2 -> O2 + H2O
Half reactions: (charge and atoms balanced)
14H+ + 6e + Cr2O7 2- -> 2Cr 3+ +7H2O
H2O2 -> O2 + H2O (The teacher told us to do it like this) For this one, how do I balance the charges and atoms? By adding H2O and H+ or just balance by coefficients?
I try it this way:
2H2O2 -> O2 + 2H2O (The H and O atoms are balanced and their charges are the same, so I have no way to add electrons) Then how can I do? By adding H2O and H+ instead of adding coefficinets?
2) Cu + HNO3 -> Cu(NO3)2 + NO +H2O (ACIDIC condition)
oxidation numbers are: 0,+1,+5,-2,->,+2,+5,-2,+2,-2,+1,-2 respectively
Cu -> Cu(NO3)2 <-this bugs me, how can I balance the N atoms
3H+ + 3e + HNO3 -> NO +2H2O
3)KIO3 + KI + HCL -> KCL + I2 + H2O (ACIDIC condition)
KIO3 -> I2 (How can I blance the K atoms with no K on the product side?)
KI -> I2 (same problem)
4)This one I am just not sure if I am right!
OF2 + I- -> F- + I3 - (ACIDIC condition)
oxidation nubmers are: +2,-1,-1,-1,-1/3 respectively
2H+ + 4e + OF2 -> 2F- + H2O (I put an H2O on the product side, beucase it is in acidic solution (the teacher told me that) and it is the only way to balance the O atoms)
2(3I- -> I3 - + 2e)
Add the half reactions:
2H+ + OF2 + 6I- -> 2F- + 2I3 - + H2O
(suprisely the charge and atoms are balanced
I feel sorry to have that many questions! (Hopefully we are done with balanceing redox equations) Please help me! Thank you a million times!