# Electrolysis of Molten Sodium Chloride ## Given Data - **Current (\(I\))**: 2 A - **Time (\(t\))**: 30 min = 180 s - **Faraday Constant (\(F\))**: 96500 C/mol - **Atomic Mass of Na**: 23 g/mol ## Step 1: Write the Cathode Reaction Molten NaCl dissociates into Na\(^+\) and Cl\(^-\). At cathode: \[ \text{Na}^+ + e^- \rightarrow \text{Na} \] So, **1 mole of Na** requires **1 mole of electrons** (1 Faraday). ## Step 2: Calculate Total Charge Passed \[ Q = I \times t = 2\,\text{A} \times 180\,\text{s} = 360\,\text{C} \] ## Step 3: Calculate Moles of Electrons (Faradays) \[ \text{Number of Faradays} = \frac{Q}{F} = \frac{360}{96500} \approx .0373\,\text{mol} \] ## Step 4: Calculate Moles of Sodium Deposited Since the reaction requires 1 mole of electrons per 1 mole of Na: \[ \text{Moles of Na deposited} = \text{Moles of electrons} \approx .0373\,\text{mol} \] ## Step 5: Calculate Mass of Sodium \[ \text{Mass of Na} = \text{Moles} \times \text{Atomic mass} \] \[ = .0373\,\text{mol} \times 23\,\text{g/mol} \] \[ = .8579\,\text{g} \] ## Final Answer **Mass of sodium deposited at the cathode:** \[ \boxed{.86\,\text{g}} \] (Rounded to two decimal places)