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a)State the Faraday’s law of electrolysis b)Calculate how long it would take an aqueous gold (III) chloride cell to coat 2.5 g of gold on a...

a)State the Faraday’s law of electrolysis

b)Calculate how long it would take an aqueous gold (III) chloride cell to coat 2.5 g of gold
on a bracelet using a current of 2.5 A. The half reaction has been provided for you.

c)Two half-cells are connected under standard conditions to make an electrochemical cell.
The two half-cells are a copper-copper (I) ion (Cu/Cu+) and an aluminum-aluminum ion (Al/Al3+). Using your the Standard Reduction Potentials below answer.

Al³⁺(aq) + 3e- → Al(s) -1.66 V
Cu²⁺(aq) + e- → Cu(s) 0.52 V

i)Write the cell diagram for the cell obtained when the two half cells are connected
ii)Identify which reaction is the anode and which is the cathode.
(iii)Calculate the emf for the cell
(iv)Write the overall balanced redox reaction for the electrochemical cell

d)An excess of copper solid is dropped into a solution which contains AgNO₃ , Fe (NO₃)3 and Zn (NO₃)2. Write the ionic equations for any reduction half cell-reactions that occur over time under standard conditions

Answers

a)The amount of substance discharged at the electrodes is directly proportional to the quantity of electricity passed.

b)(Au =197)
Au³⁺ (aq) + 3e- ------------>Au(s)

1 F=96500

3F= 96500×3=289500 C
289500=197
=2.5

(2.5 ×289500)/197=3673.86 C
Q=It=3673.86=2.5×t
t=(1469.59)/60
=24.49 mins

c)i.Al(s) // Al(aq) // Cu+(aq // Cu (s)

ii)Anode- Al(s)------------>Al³⁺ (aq) +3e-
Cathode - Cu+ + e- ------------->Cu(s)

iii)Emf= EReduced - EOxidised
0.52- -1.66= +2.18V

iv)Al(s)+ 3Cu⁺(aq -------> Al³⁺(aq) + 3Cu (s) +2.18

d)Ag⁺(aq) + e- --------------->Ag (s)

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Githiari answered the question on September 14, 2017 at 20:26

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