View Single Post
Old 02-19-2008, 12:03 AM   #2 (permalink)
n0nsensical
Junkie
 
Location: San Francisco
Note other than H2O these are all ionic compounds, K2Cr2O7 and Na2SO3 are really 2K+ + Cr2O7(2-) and 2Na+ + SO3(2-)
Write the half-reactions (looked these up in a table, not sure if you're 'supposed' to do that)
Reduction:
Cr2O7(2-) + 14H+ + 6e- -> 2Cr(3+) + 7H2O
Oxidation:
SO3(2-) + H2O -> SO4(2-) + 2H+ + 2e-

Balance the electrons
Cr2O7(2-) + 14H+ + 6e- -> 2Cr(3+) + 7H2O
3SO3(2-) + 3H2O -> 3SO4(2-) + 6H+ + 6e-

Sum
Cr2O7(2-) + 8H+ + 3SO3(2-) -> 3SO4(2-) + 2Cr(3+) + 4H2O

Get the 8H+ from 8HCl and conveniently there are also 8Cl- in the products. I hope that's everything but I'm not very good at this either. :P

Moles of SO2 in the formation of one mole I2? I don't see any SO2s or I2s. That sounds like a different problem. you need some sort of equation.

Last edited by n0nsensical; 02-19-2008 at 11:15 AM..
n0nsensical is offline  
 

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73