Chemistry Forum

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) = Cr₂O₇^2- + 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

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