Let's solve the problem step-by-step.
Step 1: Identify the given information.
- Current (I) = 0.2 amperes
- Time (t) = 1 hour
Step 2: Convert the time from hours to seconds.
[tex]\[ \text{Time (in seconds)} = 1 \text{ hour} \times 3600 \text{ seconds/hour} = 3600 \text{ seconds} \][/tex]
Step 3: Calculate the total charge (Q) passed through the solution.
Using the formula:
[tex]\[ Q = I \times t \][/tex]
[tex]\[ Q = 0.2 \text{ amperes} \times 3600 \text{ seconds} \][/tex]
[tex]\[ Q = 720 \text{ coulombs} \][/tex]
Step 4: Determine the moles of electrons transferred.
Using Faraday's constant (F), which is approximately 96485 coulombs per mole (C/mol):
[tex]\[ \text{Moles of electrons} = \frac{Q}{F} \][/tex]
[tex]\[ \text{Moles of electrons} = \frac{720 \text{ coulombs}}{96485 \text{ C/mol}} \][/tex]
[tex]\[ \text{Moles of electrons} = 0.007462 \text{ mol} \][/tex]
Step 5: Calculate the moles of copper deposited. Copper typically has a valency of 2, meaning each copper ion (Cu²⁺) requires 2 electrons for deposition.
[tex]\[ \text{Moles of copper} = \frac{\text{Moles of electrons}}{\text{Valency of copper}} \][/tex]
[tex]\[ \text{Moles of copper} = \frac{0.007462 \text{ mol}}{2} \][/tex]
[tex]\[ \text{Moles of copper} = 0.003731 \text{ mol} \][/tex]
Step 6: Determine the mass of copper deposited. Copper has an atomic mass of approximately 63.5 grams per mole.
[tex]\[ \text{Mass of copper} = \text{Moles of copper} \times \text{Atomic mass of copper} \][/tex]
[tex]\[ \text{Mass of copper} = 0.003731 \text{ mol} \times 63.5 \text{ g/mol} \][/tex]
[tex]\[ \text{Mass of copper} = 0.236928 \text{ grams} \][/tex]
Therefore, the mass of copper that is deposited is approximately 0.2369 grams.
