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Step 1: Convert the given time from minutes to seconds. $t = 10 min \times \frac{60 s}{1 min} = 600 \text{ s}$

Step 2: Calculate the total charge ($Q$) passed. $Q = I \times t$ Where $I$ is the current and $t$ is the time. $Q = 1.5 A \times 600 s = 900 \text{ C}$

Step 3: Write the half-reaction for the deposition of copper at the cathode. $\text{Cu}^{2+}(aq) + 2e^- \to Cu(s)$ This reaction shows that 2 moles of electrons are required to deposit 1 mole of copper (Cu).

Step 4: Calculate the moles of copper deposited. Faraday's constant ($F$) is $96485 C/mol e^-$. Moles of electrons ($n_e$) = $\frac{Q}{F}$ $n_e = \frac{900 C}{96485 C/mol e^-} \approx 0.009328 mol e^-$ Since 2 moles of electrons deposit 1 mole of copper, the moles of copper deposited ($n_{Cu}$) will be: $n_{Cu} = \frac{n_e}{2} = \frac{0.009328 mol e^-}{2} \approx 0.004664 \text{ mol}$

Step 5: Calculate the mass of copper deposited. The molar mass of copper is $63.5 \text{ g/mol}$. Mass ($m$) = Moles $\times$ Molar mass $m = 0.004664 mol \times 63.5 g/mol \approx 0.2964 \text{ g}$

The mass of copper deposited at the cathode is: $\boxed{0.2964 g}$

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