Calculate the time to deposit 1.17g of nickel at cathode when 5a current was passed in ni(no3)2
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Answer:
Explanation:
mass of nickel , m = 1.17g
atomic mass of nickel , M = 58.5 g/mol
so, number of mole of Ni = 1.17/58.5 = 1/50 = 0.02
so, number of electron passed = 2 × 0.2 = 0.4 [ as nickel is diatomic ]
we know, 1mole of electrons is equivalent to 96,500C Chagres.
means, 96500C charges need to deposit 1 mole of electrons,
so, amount of charge required to deposit 0.4 mole of electron = 0.4 × 96500 = 3860C
now, current is given , I = 5A
so, time required to deposit 1.17g of nickel = charge/current = 3860/5
= 772 sec or 12 min 52 second.
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