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26 June, 20:07

Consider the reaction Au (OH) 3 + Hl - Au + 12 + H2O.

Label the half-reactions as oxidation (use "O") or reduction (use "R").

21 + 2 + 2e

Aut3 + 3e → Au

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  1. 26 June, 21:19
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    Anyone know the answer?
  2. 26 June, 22:33
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    Answer: 2l - = l2 + 2e - (O)

    Au+3 + 3e - = Au (R)

    Explanation:

    I just answered the question.
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