The charge required to deposit 0.5 mol of copper from Cu 2+ solution is (F = 96500 C):

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The charge required to deposit 0.5 mol of copper from Cu2+ solution is (F = 96500 C):

Answer: 96500 C.

A 48250 C
B 24125 C
C 96500 C
D 193000 C

Correct answer: C. 96500 C

Explanation: Cu2+ + 2e- to Cu needs 2 F per mole, so 0.5 mol needs 0.5 x 2 x 96500 = 96500 C.

In a Daniell cell, electrons flow externally from the zinc anode (oxidation) through the voltmeter to the copper cathode (reduction), while the salt bridge completes the circuit.

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The charge required to deposit 0.5 mol of copper from Cu<sup>2+</sup> solution is (F = 96500 C):

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