During electrolysis, 2.70 g of copper is deposited from a copper(II) salt solution using the same quantity of electricity that deposits lead from a lead(II) salt solution. Calculate the mass of lead d

For copper (II) salt, copper ions are $\text{Cu}^{2+}$:

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

This reaction shows that 1 mole of copper metal is deposited by 2 moles of electrons.

For lead (II) salt, lead ions are $\text{Pb}^{2+}$:

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

This reaction shows that 1 mole of lead metal is deposited by 2 moles of electrons.

We use the given mass of copper and its molar mass:

$$n_{Cu} = \frac{m_{Cu}}{M_{Cu}}$$ $$n_{Cu} = \frac{2.70 \text{ g}}{63.5 \text{ g/mol}}$$ $$n_{Cu} \approx 0.04252 \text{ mol}$$

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

$$n_{e^-} = n_{Cu} \times 2$$ $$n_{e^-} = 0.04252 \text{ mol} \times 2$$ $$n_{e^-} \approx 0.08504 \text{ mol}$$

This is the quantity of electricity (in moles of electrons) that was passed through the solution.

Since the same quantity of electricity (i.e., the same moles of electrons) is passed through the lead (II) solution, we use the moles of electrons calculated in Step 3.
From the lead half-reaction, 1 mole of $\text{Pb}$ requires 2 moles of electrons.

$$n_{Pb} = \frac{n_{e^-}}{2}$$ $$n_{Pb} = \frac{0.08504 \text{ mol}}{2}$$ $$n_{Pb} \approx 0.04252 \text{ mol}$$

We use the moles of lead and its molar mass:

$$m_{Pb} = n_{Pb} \times M_{Pb}$$ $$m_{Pb} = 0.04252 \text{ mol} \times 207 \text{ g/mol}$$ $$m_{Pb} \approx 8.79164 \text{ g}$$

Rounding to three significant figures (consistent with 2.70 g):

$$m_{Pb} \approx 8.79 \text{ g}$$

The mass of lead that will be deposited is $\text{8.79 g}$.