How do I solve this redox reaction, when one of the reactants is missing (that is how my academic teacher assigned it) ?
BiO3 - + Cr3+ + (blank) = Cr2O72- + Bi 3+ + H2O
When I tried to solve it, it requiers 28 OH - in the blank space, but I think I did something wrong there. That seems just too much for me
Redox problem
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Re: Redox problem
First dichromate is not existing in alcaline solution.
So it can only be H+ what is missing.
Reduction
BiO3- + 6 H+ + 2e- => Bi 3+ + 3 H2O
Oxidation
2 Cr 3+ + 7 H2O => Cr2O7 2- + 14 H+ + 6 e-
Multiply first equation by 3
3 BiO3- + 18 H+ 6 e- => 3 Bi 3+ + 9 H2O
Addition
3 BiO3- + 18 H+ + 2 Cr 3+ + 7 H2O + 6 e- => 3 Bi 3+ + 9 H2O + 6e- + Cr2O7 2- +14 H+
Balancing
3 BiO3- + 4 H+ + 2 Cr 3+ => 3 Bi 3+ + 2 H2O + Cr2O7 2-
Finished
So it can only be H+ what is missing.
Reduction
BiO3- + 6 H+ + 2e- => Bi 3+ + 3 H2O
Oxidation
2 Cr 3+ + 7 H2O => Cr2O7 2- + 14 H+ + 6 e-
Multiply first equation by 3
3 BiO3- + 18 H+ 6 e- => 3 Bi 3+ + 9 H2O
Addition
3 BiO3- + 18 H+ + 2 Cr 3+ + 7 H2O + 6 e- => 3 Bi 3+ + 9 H2O + 6e- + Cr2O7 2- +14 H+
Balancing
3 BiO3- + 4 H+ + 2 Cr 3+ => 3 Bi 3+ + 2 H2O + Cr2O7 2-
Finished