According to Faraday's First Law of Electrolysis, the mass of a substance deposited at an electrode is given by the formula:

$m = Z \times Q = Z \times I \times t$

Where: $Z = 30 \times 10^{-5} \text{ g C}^{-1}$ (electrochemical equivalent) $I = 1.5 \text{ A}$ (current) $t = 10 \text{ min} = 10 \times 60 \text{ s} = 600 \text{ s}$ (time)

First, calculate the total charge ($Q$): $Q = I \times t = 1.5 \text{ A} \times 600 \text{ s} = 900 \text{ C}$

Now, calculate the mass ($m$) deposited: $m = 30 \times 10^{-5} \text{ g C}^{-1} \times 900 \text{ C}$ $m = 27000 \times 10^{-5} \text{ g} = 0.27 \text{ g}$

Therefore, 0.27 g of copper is deposited on the electrode.