What number of moles of Cu 2+ will be deposited by 360 coulombs of electricity?
[f = 96500 C mol -1]
5.36 x 10 -4 mole
1. 87 x 10 10 -3 mole
9. 35 x 10 -4 mole
3. 73 x 10 -3
Correct answer is B
Cu\(^{2+}\) + 2e\(^{-}\)\(_{(aq)}\) \(\rightarrow\) Cu\(_{(s)}\)
Liberating 1 mole of Cu
2 x 1F = 2 x 96,500C = 193,000C of electricity is needed
Hence, 193,000C = 1 mole
360C of electricity = \(\frac{360C \times 1 mol}{193,000C}\)
= 0.001865 mol
= 1.87 x 10\(^{-3}\) moles