Answer:
A. Oxidation refers to loss of electron or increase in oxidation state. Reduction refers to gain of electrons or decrease in oxidation state.
B. The oxidation number of S in S₂O₃²⁻ is +2.
C. The oxidation number of S in S₄O₆²⁻ is +5/2.
Explanation:
A. Oxidation refers to loss of electron or increase in oxidation state. Reduction refers to gain of electrons or decrease in oxidation state.
The overall charge on a molecule is equal to the sum of the oxidation states of all the atoms in a molecule.
B. S₂O₃²⁻ : The oxidation state of oxygen (O) is -2 and overall charge on molecule is -2.
Therefore, the oxidation state of sulfur (x) can be calculated by
2x + 3 (-2) = -2
2x + (-6) = -2
2x = -2 + 6 = 4
x = 4 ÷ 2= +2
<u>Therefore, the oxidation number of S is +2.</u>
C. S₄O₆²⁻ : The oxidation state of oxygen (O) is -2 and overall charge on molecule is -2.
Therefore, the oxidation state of sulfur (x) can be calculated by
4x + 6 (-2) = -2
4x + (-12) = -2
4x = -2 + 12 = 10
x = 10 ÷ 4 = +5/2
<u>Therefore, the oxidation number of S is +5/2.</u>
