What number of moles of Cu 2+ will be deposited by 360 coulombs of electricity?
[f = 96500 C mol -1]
[f = 96500 C mol -1]
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Correct Answer: Option B
Explanation:
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
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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
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