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15 March, 07:29

A current of 0.44 A is passed through a solution of a ruthenium nitrate salt, causing reduction of the metal ion to the metal. After 25.0 minutes, 0.345 g of Ru (s) has been deposited. What is the oxidation state of ruthenium in the nitrate salt?

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  1. 15 March, 09:34
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    Ruthenium has an oxidation state of + 2

    Explanation:

    Step 1: Data given

    A current = 0.44 A

    Time = 25.0 minutes

    Mass of Ru (s) = 0.345 grams

    Step 2: Calculate charge

    Charge = current * time

    Charge = 0.44 A * 25 min * 60 sec/min

    Charge = 660 C

    Step 3: Calculate moles electrons

    1 mol e - = 660 C * (1 mol e- / 96500 C)

    1 mol e - = 6.84 * 10^-3 mol e-

    Step 4: Calculate moles of Ru

    Moles Ru = mass Ru / molar mass Ru

    Moles Ru = 0.345 grams / 101 g/mol

    Moles Ru = 3.416 * 10^-3 moles Ru

    Step 5: Calculate oxidation state of ruthenium

    (6.84e-3 moles e-) / (3.416e-3 moles Ru) = 2 e - / Ru atom (Ru+2)

    Ruthenium has an oxidation state of + 2

    Since a nitrate ion has a charge of - 1, ruthenium nitrate would be

    Ru (NO3) 2
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