What mass of zinc will be deposited from A solution of zinc 2 chloride when a current of 3A Is passed through the zinc (ii) chloride Solution during electrolysis of 50 minutes(Zn=65 , 1Faraday=96500C)

The given time is $50 \text{ minutes}$.

$$t = 50 \text{ min} \times 60 \text{ s/min}$$ $$t = 3000 \text{ s}$$

The given current is $I = 3 \text{ A}$.

$$Q = I \times t$$ $$Q = 3 \text{ A} \times 3000 \text{ s}$$ $$Q = 9000 \text{ C}$$

Zinc(II) ions ($Zn^{2+}$) gain electrons to form solid zinc.

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

This reaction shows that $2 \text{ moles of electrons}$ are required to deposit $1 \text{ mole of Zn}$.

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

$$\text{Moles of electrons} = \frac{\text{Charge (Q)}}{\text{Faraday's constant (F)}}$$ $$\text{Moles of electrons} = \frac{9000 \text{ C}}{96500 \text{ C/mol e}^-}$$ $$\text{Moles of electrons} \approx 0.093264 \text{ mol e}^-$$

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

$$\text{Moles of Zn} = \frac{\text{Moles of electrons}}{2}$$ $$\text{Moles of Zn} = \frac{0.093264 \text{ mol}}{2}$$ $$\text{Moles of Zn} \approx 0.046632 \text{ mol}$$

The molar mass of Zn is $65 \text{ g/mol}$.

$$\text{Mass of Zn} = \text{Moles of Zn} \times \text{Molar mass of Zn}$$ $$\text{Mass of Zn} = 0.046632 \text{ mol} \times 65 \text{ g/mol}$$ $$\text{Mass of Zn} \approx 3.031 \text{ g}$$

The mass of zinc deposited is approximately $\text{3.031 g}$.