For the equation below, write the oxidation half-reaction and the reduction half-reaction. 2Na + Cl2 =–> 2NaCl

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For the equation below, write the oxidation half-reaction and the reduction half-reaction. 2Na + Cl2 —–> 2NaCl

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Sigridomena 3 years 2021-08-21T16:05:03+00:00 1 Answers 4 views 0

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  1. Farah
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    2021-08-21T16:06:06+00:00
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    Answer:

    Oxidation half-reaction : Na(s) → Na⁺ + 1 e-

    Reduction half-reaction: Cl₂ + 2 e- → 2 Cl⁻

    Explanation:

    Oxidation half-reaction: solid sodium (Na(s)) has an oxidation number of 0. It loses 1 electron and forms the cation Na⁺. So, the half-reaction is the following:

    Na(s) → Na⁺ + 1 e-

    Reduction half-reaction: chlorine gas (Cl₂) has an oxidation number of 0. Each atom of Cl gains 1 electron to form two Cl⁻ ions, according to the following half-reaction:

    Cl₂ + 2 e- → 2 Cl⁻

    The total oxidation-reduction reaction is obtained by adding the oxidation half-reaction multiplied by 2 (to balance the electrons) and the reduction half-reaction, as follows:

    2 x (Na(s) → Na⁺ + 1 e-)

    Cl₂ + 2 e- → 2 Cl⁻

    ——————————–

    2Na(s) + Cl₂ → 2NaCl

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )