If the same quantity of electricity is passed through two electrolytic cells, one containing copper(II) sulfate solution and the other containing lead(II) nitrate solution, how much lead will be depos

For Copper (Cu):

• Mass of copper ($m_{Cu}$) = 2.70 g
• Molar mass of copper ($mm_{Cu}$) = 63.5 g/mol
• Number of electrons transferred for copper (II) ($n_{Cu}$): The half-reaction is $\text{Cu}^{2+}(aq) + 2e^- \to \text{Cu}(s)$, so $n_{Cu} = 2$.

For Lead (Pb):

• Molar mass of lead ($mm_{Pb}$) = 207 g/mol
• Number of electrons transferred for lead (II) ($n_{Pb}$): The half-reaction is $\text{Pb}^{2+}(aq) + 2e^- \to \text{Pb}(s)$, so $n_{Pb} = 2$.
• Mass of lead ($m_{Pb}$) = ? (This is what we need to calculate)

Since $n_{Cu} = n_{Pb} = 2$, we can cancel out the '2' from both sides of the equation:

$$\frac{2.70}{63.5} = \frac{m_{Pb}}{207}$$

Now, isolate $m_{Pb}$:

$$m_{Pb} = \frac{2.70}{63.5} \times 207$$ $$m_{Pb} = 0.042519685 \times 207$$ $$m_{Pb} = 8.7965748 \text{ g}$$

$$m_{Pb} \approx 8.80 \text{ g}$$

The mass of lead that will be deposited is $\text{8.80 g}$.