A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 ampere. The mass of copper deposited at the cathode is:

Molar mass of Cu=63 g mol⁻¹, 1F =96487 C mol⁻¹

**Step 1: Identify the electrochemical reaction**
At the cathode copper(II) ions are reduced to metallic copper:

$Cu^\{2+\}+2e^\{-\}\backslash rightarrow\; Cu\backslash ;(s)$

**Step 2: Convert the electrolysis time to seconds**
The current is applied for 10 minutes.

$t\; =\; 10\backslash \; \backslash text\{min\}\backslash times\; 60\backslash \; \backslash frac\{\backslash text\{s\}\}\{\backslash text\{min\}\}\; =\; 600\backslash \; \backslash text\{s\}$

**Step 3: Calculate the total charge passed**
Charge (Q) equals current (I) multiplied by time (t).

$Q\; =\; I\; \backslash times\; t\; =\; 1.5\backslash \; \backslash text\{A\}\backslash times\; 600\backslash \; \backslash text\{s\}=\; 900\backslash \; \backslash text\{C\}$

**Step 4: Determine the amount of copper deposited (in moles)**
Using Faraday’s law:

$n\_\{\backslash text\{Cu\}\}\; =\; \backslash frac\{Q\}\{n\backslash ,F\}\; =\; \backslash frac\{900\backslash \; \backslash text\{C\}\}\{2\; \backslash times\; 96487\backslash \; \backslash text\{C\; mol^\{-1\}\}\}$

$n\_\{\backslash text\{Cu\}\}\; \backslash approx\; \backslash frac\{900\}\{192974\}\backslash \; \backslash text\{mol\}\backslash approx\; 4.665\backslash times10^\{-3\}\backslash \; \backslash text\{mol\}$

**Step 5: Convert moles of copper to mass**
Molar mass of Cu = 63 g mol⁻¹.

$m\; =\; n\_\{\backslash text\{Cu\}\}\backslash times\; M\; =\; 4.665\backslash times10^\{-3\}\backslash \; \backslash text\{mol\}\backslash times\; 63\backslash \; \backslash frac\{\backslash text\{g\}\}\{\backslash text\{mol\}\}$

$m\; \backslash approx\; 0.2938\backslash \; \backslash text\{g\}$

**Result**
The mass of copper deposited at the cathode after 10 minutes with a 1.5 A current is **0.2938 g**.