How do you balance redox reactions!!!
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How do you balance redox reactions!!!

[From: ] [author: ] [Date: 11-09-09] [Hit: ]
So you must use H+ on one side and H2O on the other to balance the equation.(On another occasion, you will balance equations under basic conditions. Then, you will use OH- on one side and H2O on the other to balance.)There are two ways to balance redox reactions: Electron-transfer or half-reactions.......
The oxidation number of an element in its elemental state is 0.
The oxidation numbers of all the elements in a neutral compound add up to 0.
The oxidation numbers of all the elements in an ion add up to the charge on the ion.

PH3 + I2 ===> H3PO2 + I-

This is already an advanced state of balancing redox equations. There are two conditions, acid and basic. Because one of the compounds is H3PO2, an acid, you know that these are acid conditions. So you must use H+ on one side and H2O on the other to balance the equation.

(On another occasion, you will balance equations under basic conditions. Then, you will use OH- on one side and H2O on the other to balance.)

There are two ways to balance redox reactions: Electron-transfer or half-reactions. It is hard to show electron-transfer on a computer, so let's do half-reactions. In PH3, H = -1, so P = +3. In H3PO2-, H = 1, and O = -2, so P = 0. So P is reduced from +3 to 0. Using H+'s and H2O's this leads us to the first half-reaction:

(1) 2H2O + PH3 ===> H3PO2- + 3e- + 4H+

Balancing two I atoms on each side, the second half reaction must be:

(2) I2 + 2e- ===> 2I-

Electrons lost must equal electrons gained, so multiply (1) by 2 and multiply (2) by 3:

(3) 4H2O + 2PH3 ===> 2H3PO2- + 6e- + 8H+
(4) 3I2 +6e- ===> 6I-

Add (3) and 4) together and cancel on each side:

4H2O 2PH3 + 3I2 ===> 2H3PO2 + 6I- + 8H+
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