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Calculate the time required to deposit 56g of silver from a silver nitrate solution using a current of 4.5A.
n(e) = 0.519 mol F = 96,500 C mol-1Q = 0.519 x 96,500 = 50,083.5 CQ = 50,083.5 C I = 4.5 A
t = Q/It = 50,083.5 ÷ 4.5 = 11,129.67 seconds t = 11,129.67 ÷ 60 = 185.5 minutes t = 185.5 ÷ 60 = 3.1 hours
WIEGHT DEPOSITED=EQUIVALENT WIEGHT*CURRENT(i)*TIME TAKEN FOR THE DEPOSIT(t)/96500 AND THUS 45=108*4.5*t/96500
WIEGHT DEPOSITED=EQUIVALENT WIEGHT*CURRENT(i)*TIME TAKEN FOR THE DEPOSIT(t)/96500
AND THUS 45=108*4.5*t/96500
Mass of Ag= 56 gmMolar mass of Ag= 107.9 gm/mol Number of moles of Ag= mass÷molar mass = 56÷107.9 = 0.518999 Ag+ + e-= Ag To deposite 1 mole Ag, 1 mole e- is passed So, to deposite 0.518999 mole, Number of e- passed=0.518999×1=0.518999 moles •1 mole consist of 96500C So, 0.518999 mole charge contain is, Amount of charge passed= 96500×0.518999 =50083.4035C Time taken = charge÷current = 50083.4035÷4.5 time = 11129.645 second time= 1129.645÷60=185.494 minutes time = 185.494÷60=3.092 hour
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