Step 5: Balance the hydrogen atoms by adding H+ ions where needed.
7 H’s on left, 3H’s on right, add 4 H+ on right
2 H2O + PH3 → H3PO2 + 4 H+
Step 6: Balance the charge by adding electrons, e-.
2 H2O + PH3 → H3PO2 + 4 H+ + 4 e-
2 e- + I2 → 2 I-
Step 7: If the number of electrons lost in the oxidation half-reaction is not equal to the number of electrons gained in the reduction half-reaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number of electrons lost.
Multiply bottom equation by 2
4 e- + 2 I2 → 4 I-
Step 8: Add the 2 half-reactions as if they were mathematical equations. The electrons will always cancel. If the same formulas are found on opposite sides of the half-reactions, you can cancel them. If the same formulas are found on the same side of both half-reactions, combine them.
2 H2O + PH3 + 2 I2 → H3PO2 + 4 H+ + 4 I-1
Combine 4 H+ and 4 I - = 4 HI
2 H2O + PH3 + 2 I2 → H3PO2 + 4HI
Check
4 H’s + 3 H’s = 3 H’s + 4 H’s
2 O’s = 2 O’s
1 P = 1P
4 I’s = 4 I’s
OK
IF the H3PO2 is -1
Step 6: Balance the charge by adding electrons, e-.
2 H2O + PH3 → H3PO2^-1 + 4 H+ + 3 e-
2 e- + I2 → 2 I-
Step 7:
Multiply top equation by 2 and bottom by 3 to have 6 electrons.
4 H2O + 2 PH3 → 2 H3PO2^-1 + 8 H+ + 6 e-
6 e- + 3 I2 → 3 I-
Step #8 add’em up
4 H2O + 2 PH3 + 3 I2 → 2 H3PO2^-1 + 8 H+ + 6 I-
For basic solution:
Convert all H+ to H2O. Do this by adding OH¯ ions to both sides. The side with the H+ will determine how many hydroxide ions to add.
I hope this helps!