Elle_Mendelson_2K wrote:Will someone explain how to balance d in chapter 14 problem 5 in the 6th edition. I don't understand why the OH is on the right side instead of the left when you would add H2O on the left and therefore OH- on the right
First, you want to balance the oxygen by adding 8 H2O molecules on the left. Now, balance the hydrogen atoms by adding 8H2O and 8OH- on the right and left side respectively. Now the 8H2Os cancel and leaves the equation as:
P4(s) + 8OH- (aq) --> 4H2PO2- (aq)
Finally just add 4e- to the right to balance the charges to get your balanced oxidation half reaction.
For the reduction half-reaction, you're going to add 12H2O on the left and 12OH- on the fight, and 12e- on the left to balance the charge. This gives you:
P4(s)+12H2O+12e- --> 4PH3(g)+12OH-
Cancel like terms and multiply the reactions by constants to find final reaction which is:
P4(s)+3H2O+3OH- ---> 3H2PO2- + PH3(g)