Cr(OH)3+IO3^-1 =Cr^+3+I^-1 Balance the reaction by half reaction method
Answers
Answer:
Its been 30 years since I have done this… I doubt things have changed. Split the reaction into 2 parts (each is a reduction OR oxidation).
IO3(-1) → I(-1) and Cr(OH)3 → CrO4(-2)
determine electrons lost and gained…
I(+5) to I(-1) so 6 electrons 6(-) + IO3(-1) → I(-1)
Cr(+3) to Cr(+6) so 3 electrons Cr(OH)3 → CrO4(-2) + 3e(-)
add water and OH-/H+ as needed to balance O’s
6H(+) + 6e(-) + IO3(-1) → I(-1) +3H2O
H2O + Cr(OH)3 → CrO4(-2) + 3e(-) + 5H(+)
multiple something to have the same number of electrons on both sides… multiply Cr half reaction by 2…
2H2O + 2Cr(OH)3 → 2CrO4(-2) + 6e(-) + 10H(+)
and subtract the half equations…
2H2O + 2Cr(OH)3 → 2CrO4(-2) + 6e(-) + 10H(+)
6H(+) + 6e(-) + IO3(-1) → I(-1) +3H2O
equals = 2H2O + 2Cr(OH)3 + IO3(-1) + 6H(+) → 2CrO4(-2) + I(-1) + 10H(+) + 3H2O
then reduce any items on both sides of arrows… H2Os and H(+)s
2Cr(OH)3 + IO3(-1) → 2CrO4(-2) + I(-1) + 4H(+) + 1H2O
make it a bit more pretty by balancing the charges on both sides by adding Hs —- I added 1H(+) to both sides to make chemistry pretty…
2Cr(OH)3 + HIO3 → 2H2CrO4 + HI + H2O
phew lots of work.
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2Cr(OH)3 + 2H2O –> 2CrO4(-2) +10H(+)+6e-
IO3(-) + 6H(+) + 6e- –> I(-) + 3H2O
net:
2Cr(OH)3+IO3(-1)
-> I(-1)+2CrO4(-2)+H2O+4H(+)
based on KIO3:
2Cr(OH)3 + KIO3
-> KI + 2H2CrO4 + H2O
hope it's helpful
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