Question
Assuming that copper contains only iron, silver and gold as impurities. After passage of 12.4 A for 4825 s, the mass of anode decreased by 20.00 g and the cathode increased by 19.05 g. The percentages of iron and copper in the original sample are, respectively, (Cu = 63.5, Fe = 56)
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Solution
neq Cu deposited at cathode = Q/F
Or, $\frac{w}{63.5} \times 2 = \frac{12.4 \times 4825}{96500} \Rightarrow w = 19.685\text{ gm}$
But the increase in mass of cathode is only 19.05 gm It represents that 20 gm of sample contains only 19.05 gm Cu.
$\therefore\%\text{ of Cu = }\frac{19.05}{20} \times 100 = 95.25\%$
Now, $\frac{Q}{F} = n_{eq}Cu + n_{eq}Fe\left( \text{Oxidised at Anode} \right)$
Or, $\frac{12.4 \times 4825}{96500} = \frac{19.05}{63.5} \times 2 + \frac{w}{56} \times 2 \Rightarrow w = 0.56\text{ gm}$
∴ Percentage of Fe = $\frac{0.56}{20} \times 100 = 2.8\%$
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