Question

Calculate current in ampere required to deposit 4.8 g Cu from it's salt solution in 30 minutes. [Molar mass of $Cu=63.5\text{ g mol}^{-1}$]

• A. 8.1 ampere
• B. 6.4 ampere
• C. 10.5 ampere
• D. 12.3 ampere

Hint:

Logic Tip: Always convert time strictly to seconds when using Amperes and Faraday's constant, because $1\text{ Ampere} = 1\text{ Coulomb / second}$. Forgetting to multiply 30 minutes by 60 is the most common pitfall in these calculations!

Answer 1

The Correct Option is A

Concept:
According to Faraday's First Law of Electrolysis, the mass ($W$) of a substance deposited at an electrode is directly proportional to the total electric charge ($Q$) passed through the solution. $$W = Z \cdot I \cdot t$$ where $I$ is the current in Amperes, $t$ is the time in seconds, and $Z$ is the electrochemical equivalent ($Z = \frac{\text{Molar Mass{nF}$).
Step 1: Determine the number of electrons (n) exchanged.
Copper in a standard aqueous salt solution (like $CuSO_4$) exists as a $Cu^{2+}$ ion. The reduction reaction at the cathode is: $$Cu^{2+}_{(aq)} + 2e^- \longrightarrow Cu_{(s)}$$ This means 2 moles of electrons are required to deposit 1 mole of copper. So, $n = 2$.
Step 2: Set up Faraday's formula with the given values.
Given data: Mass deposited, $W = 4.8\text{ g}$ Time, $t = 30\text{ minutes} = 30 \times 60 = 1800\text{ seconds}$ Molar Mass of Cu, $M = 63.5\text{ g/mol}$ Faraday's Constant, $F = 96500\text{ C/mol}$ Substitute these into the extended formula $W = \frac{M}{nF} \times I \times t$: $$4.8 = \frac{63.5}{2 \times 96500} \times I \times 1800$$
Step 3: Solve for the current I.
Rearrange the equation to isolate $I$: $$I = \frac{4.8 \times 2 \times 96500}{63.5 \times 1800}$$ $$I = \frac{9.6 \times 96500}{114300}$$ $$I = \frac{926400}{114300}$$ $$I \approx 8.105\text{ A}$$ Rounding to one decimal place, the required current is $8.1\text{ A}$.