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6 June, 09:16

A. Define the terms 'oxidation' and 'reduction.

B. The oxidation number of S

inS2O32 - is:

C. The oxidation number of S

inS4O62 - is:

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  1. 6 June, 10:21
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    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

    Therefore, the oxidation number of S is + 2.

    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

    Therefore, the oxidation number of S is + 5/2.
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