If 3.50 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic...
If 3.50 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.83 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge. (answer in mA)
Homework Answers
3.50 × 10-3 Kg gold = 0.018 mole gold
Total charge in this amount of gold ion is
Q = 0.018× 6.022×1023×1.6×10-19 C
= 1.73 × 102 C
I = Q ÷ t
= { 1.73 ×102 ÷ (2.83×3600)} amp
= 0.01698 amp
= 16.98 mA
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