During Purification of copper By electrolysis 1.48 grams of copper were deposited when A current was passed through Ikea's copper 2 sulfate for two and a half hours Calculate the amount of current passed(Cu=63.5 1Faraday=96500C)

During electrolysis, copper(II) ions gain electrons to form solid copper.

$$Cu^{2+}(aq) + 2e^- \to Cu(s)$$

The molar mass of copper (Cu) is $63.5 \text{ g/mol}$.

$$\text{Moles of Cu} = \frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}$$ $$\text{Moles of Cu} = \frac{1.48 \text{ g}}{63.5 \text{ g/mol}}$$ $$\text{Moles of Cu} \approx 0.023307 \text{ mol}$$

From the half-reaction, $1 \text{ mole of Cu}$ requires $2 \text{ moles of electrons}$.

$$\text{Moles of electrons} = \text{Moles of Cu} \times 2$$ $$\text{Moles of electrons} = 0.023307 \text{ mol} \times 2$$ $$\text{Moles of electrons} \approx 0.046614 \text{ mol}$$

Faraday's constant is $96500 \text{ C/mol e}^-$.

$$\text{Charge (Q)} = \text{Moles of electrons} \times \text{Faraday's constant}$$ $$\text{Charge (Q)} = 0.046614 \text{ mol} \times 96500 \text{ C/mol}$$ $$\text{Charge (Q)} \approx 4500.85 \text{ C}$$

The time given is $2.5 \text{ hours}$.

$$\text{Time (t)} = 2.5 \text{ hours} \times 60 \text{ min/hour} \times 60 \text{ s/min}$$ $$\text{Time (t)} = 9000 \text{ s}$$

The relationship between charge, current, and time is $Q = I \times t$.

$$I = \frac{Q}{t}$$ $$I = \frac{4500.85 \text{ C}}{9000 \text{ s}}$$ $$I \approx 0.500 \text{ A}$$

The amount of current passed is approximately $\text{0.500 A}$.