Question

__________ electrons appear in the following half-reaction when it is balanced.

S₄O₆^2-→ 2S₂O₃^2-

Select one:

a. 1

b. 4

c. 6

d. 3

e. 2

Answer 1

Step 1:

Oxidation number of O = -2

lets the oxidation number of S be x

use:

6* oxidation number (O) + 4* oxidation number (S) = net charge

6*(-2)+4* x = -2

-12 + 4 * x = -2

4x = 10

x = 2.5

So oxidation number of S = +2.5

Oxidation number of each element in S4O6-2 is

O=-2

S=+2.5

Step 2:

Oxidation number of O = -2

lets the oxidation number of S be x

use:

3* oxidation number (O) + 2* oxidation number (S) = net charge

3*(-2)+2* x = -2

-6 + 2 * x = -2

x = 2

So oxidation number of S = +2

Oxidation number of each element in S2O3-2 is

O=-2

S=+2

Step 3:

Oxidation state of S changes from 2.5 to 2

There are 4 atom of S taking part in reaction.

1 atom of S needs 1/2 electron

So, 4 atoms would need 4*(1/2) = 2 electrons

Answer: e