ElectrochemistryHard

Question

The current efficiency of an electrode position of copper metal in which 9.8 g of copper is deposited by a current of 3 A for 10000 s, from aqueous copper sulphate solution, is about

Options

A.60%
B.99%
C.92%
D.75%

Solution

$\frac{W}{E} = \frac{Q}{F} \Rightarrow \frac{9.8}{63.5} \times 2 = \frac{3 \times 10000}{96500} \times \eta \Rightarrow \eta = 0.99$

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