Hello, I am trying to balance a tricky (skeleton) reaction using the half reaction method: CrI3 (s) + H2O2 (aq) → CrO42- (aq) + IO4- (aq) I ended up using H2O to balance the peroxide for the reduction half and pulled the rest of the equation down for the oxidation half since Cr and I both lose electrons. The final result was: 27H202 + 2CrI3 -----> 2CrO4^2- +6IO4^- + 22H2O + 10H+ for acidic  and... 27H202 + 2CrI3 +10 OH^- -----> 2CrO4^2- +6IO4^- + 32H2O for basic Atoms and charges look balanced but I am feeling a bit uncertain about my method. Can you please confirm whether this is correct?  Thank you, Carolyn

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Hello,

I am trying to balance a tricky (skeleton) reaction using the half reaction method:

CrI3 (s) + H2O2 (aq) → CrO42- (aq) + IO4- (aq)

I ended up using H2O to balance the peroxide for the reduction half and pulled the rest of the equation down for the oxidation half since Cr and I both lose electrons. The final result was:

27H202 + 2CrI3 -----> 2CrO4^2- +6IO4^- + 22H2O + 10H+ for acidic 

and...

27H202 + 2CrI3 +10 OH^- -----> 2CrO4^2- +6IO4^- + 32H2O for basic

Atoms and charges look balanced but I am feeling a bit uncertain about my method. Can you please confirm whether this is correct? 

Thank you,

Carolyn

 

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