Redox equations in Chemistry
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Redox equations in Chemistry

[From: ] [author: ] [Date: 11-07-01] [Hit: ]
2 S’s = +4, S = +2+2 + (2 * +2) + (3 * -2) = +2 + +4 + -6 = 0In Na2S4O6, 2 Na’s = +2, S4O6 = -2 6 O’s = -12, so 4 S’s = +10S = +2.5+2 + (4 * 2.......


(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.

-
The first clue is the formation of pure elements. This is a reduction reaction. If the pure element is turned to a salt, it is oxidized. Oxidized compounds increase their charge and reduced ones reduce it.

(Simple) example

Fe + 2HCl -> FeCL2 + H2

It helps to be able to write the ionic equation first.

Fe + 2H(+) + 2Cl(-) -> Fe(+2) + 2Cl(-) + H2

The net ionic equation subtracts all ions common to both sides of the equation.

Fe + 2H(+) -> Fe(+2) + H2

Now it is rather simple to see that the Iron has been oxidized and the Hydrogen reduced.

Not so (simple) example

HCl(aq) + HNO3(aq) + Sn(s) → NO2(g) + H2SnCl6(aq) + H2O(l)

First, get rid of the silly state subscripts. Actual working chemists never use them.
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