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Discussion Response Post to: Electrochemistry..

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Srujana (2785)

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Quantity of electricity required to deposite 1 mol of a substance is given by Q=nF (n is the oxidation number)

(1.17)(25*60+15) C is required to produce 0.583 g of copper,

nF produces 1 mole of copper = 63.54 g

$n=\frac{63.54 \times 1.17 \times 1515}{0.583\times 96500}$

which gives n=2

copper will be deposited at cathode when it gains electrons,hence this is a reduction reaction and the oxidation state of copper is +2

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