Hydrogen peroxide will oxidize iodide to iodine: (sulfuric acid is used to provide a source of hydrogen ions to neutralize the KOH that is formed)
H2O2 + 2 KI ---> 2KOH + I2
2KOH + H2SO4 ---> K2SO4 + 2 H2O
An alternative reaction could also be written:
2 KI + H2SO4 ---> 2HI + K2SO4
2HI + H2O2 ---> I2 + 2 H2O
Iodine will oxidize thiosulfate to sulfate:
S2O3-- ---> SO4--
To balance the equation, need to find the oxidation states of S in thiosulfate and sulfate:
S2O3-- has 3O-- or 6- from the oxygen, subtract 2- for the ion leaves 4-, therfore each S must be +2 (to leave 2 neg charges for the ion).
SO4-- has 4O-- or 8- from the oxygen, subtract 2- for the ion leaves 6-, therefore each S must be +6 (to leave 2 neg charges for the ion)
So S+2 ---> S+6 + 4e-
or 2 (S+2) ---> 2(S+6 ) + 8e-
each iodine atom needs one electron to form I-
4 I2 + 8e- ---> 8I-
So now we can write a balanced equation:
4 I2 + Na2S2O3 + 5 H2O ---> 2 NaHSO4 + 8HI
H2O2 + 2 KI ---> 2KOH + I2
2KOH + H2SO4 ---> K2SO4 + 2 H2O
An alternative reaction could also be written:
2 KI + H2SO4 ---> 2HI + K2SO4
2HI + H2O2 ---> I2 + 2 H2O
Iodine will oxidize thiosulfate to sulfate:
S2O3-- ---> SO4--
To balance the equation, need to find the oxidation states of S in thiosulfate and sulfate:
S2O3-- has 3O-- or 6- from the oxygen, subtract 2- for the ion leaves 4-, therfore each S must be +2 (to leave 2 neg charges for the ion).
SO4-- has 4O-- or 8- from the oxygen, subtract 2- for the ion leaves 6-, therefore each S must be +6 (to leave 2 neg charges for the ion)
So S+2 ---> S+6 + 4e-
or 2 (S+2) ---> 2(S+6 ) + 8e-
each iodine atom needs one electron to form I-
4 I2 + 8e- ---> 8I-
So now we can write a balanced equation:
4 I2 + Na2S2O3 + 5 H2O ---> 2 NaHSO4 + 8HI
