(d) I2(s) + Na2S2O3(aq) → Na2S4O6(aq) + NaI(aq)
-oxidizing & reducing agents
I 0 is reduced to I-1
In Na2S2O3, 2 Na’s = +2,
S2O3 = -2, 3 O’s = -6, 2 S’s = +4, S = +2
+2 + (2 * +2) + (3 * -2) = +2 + +4 + -6 = 0
In Na2S4O6, 2 Na’s = +2,
S4O6 = -2
6 O’s = -12, so 4 S’s = +10
S = +2.5
+2 + (4 * 2.5) + (6 * -2) = +2 + 10 + -12 = 0
S+2 is oxidized to S+2.5, as I 0 is reduced to I-1
I2 is the oxidizing agent and Na2S2O3 is the reducing agent.
(d) I2(s) + Na2S2O3(aq) → Na2S4O6(aq) + NaI(aq)
I2 → 2 NaI
Balance Na+1’s by adding 2 Na+1 on left side
2 Na+1 + I2 → 2 NaI
balance charge by adding 2 e- to left side
2 e- + 2 Na+1 + I2 → 2 NaI
Na2S2O3 → Na2S4O6
Balance S’s, by making 2 Na2S2O3 on left side
2 Na2S2O3 → Na2S4O6
Balance Na+1’s by adding 2Na+1 to the right side
2 Na2S2O3 → Na2S4O6 + 2 Na+1
balance charge by adding 2 e- to right side
2 Na2S2O3 → Na2S4O6 + 2 Na+1 + 2e-
Add the 2 equations
2 e- + 2 Na+1 + I2 + 2 Na2S2O3 → 2 NaI + Na2S4O6 + 2 Na+1 + 2e-
subtract 2e- and 2 Na+1 from both sides
I2 + 2 Na2S2O3 → 2 NaI + Na2S4O6
This is balanced redox equation.