Q.A solution of $CuSO_4$ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

The key idea is Faraday's laws of electrolysis: the mass of a substance deposited is proportional to the quantity of charge passed.

Step 1 — Find total charge passed.

Current $I = 1.5 \, \text{A}$, time $t = 10 \, \text{min} = 600 \, \text{s}$.

Charge $Q = I \times t = 1.5 \times 600 = 900 \, \text{C}$.

Step 2 — Determine moles of electrons.

Faraday's constant $F = 96487 \, \text{C mol}^{-1}$.

Moles of electrons $= \frac{Q}{F} = \frac{900}{96487} \approx 0.00933 \, \text{mol}$.

Step 3 — Relate electrons to copper deposited.

Cathode reaction: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$.

So 2 moles of electrons deposit 1 mole of Cu.

Moles of Cu $= \frac{0.00933}{2} \approx 0.004664 \, \text{mol}$.

Step 4 — Convert to mass.

Using the molar mass of Cu as $63 \, \text{g mol}^{-1}$ (the value NCERT uses here): …

The key to this problem is that electrolysis forces a redox reaction using electricity. When you pass a current through a copper sulfate solution, the copper ions ($Cu^{2+}$) are attracted to the negative electrode (the cathode), where each ion gains two electrons and plates out as neutral copper metal.

The question is: how much copper? Faraday's laws give the exact link between the charge passed and the amount of substance produced.

Faraday's First Law of Electrolysis states that the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (charge, $Q$) passed through the electrolyte.

The constant of proportionality is the electrochemical equivalent ($Z$), so $m = Z \cdot Q$, where $Z$ is the mass deposited per unit charge. We can find $Z$ from the molar mass and Faraday's constant.

Step-by-step solution

1. Find the total charge passed.

Current is the rate of flow of charge: $I = Q/t$, so $Q = I \times t$ (time in seconds).

2. Determine the electrochemical equivalent of copper.

where $F$ is Faraday's constant ($96487 \text{ C mol}^{-1}$). For copper in $CuSO_4$, the ion is $Cu^{2+}$ (valency 2) and the atomic mass used here is $63 \text{ g mol}^{-1}$:

3. Apply Faraday's first law.

Method: Faraday's Laws of Electrolysis (First Law)

Concept first:

Faraday's First Law states that the mass of a substance deposited (or liberated) at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.

where $Q = I \times t$ (charge in coulombs), $I$ = current in amperes, $t$ = time in seconds.

The proportionality constant is the electrochemical equivalent ($Z$), so:

where $M$ = molar mass (g/mol), $n$ = electrons transferred per ion, $F$ = Faraday's constant.

Steps for this problem

1. Identify the cathode half-reaction

2. Write the formula for mass deposited

3. Insert known values

   • $M$ of copper = 63 g/mol (the value NCERT uses here)
   • $I = 1.5 \, \text{A}$
   • $t = 10 \, \text{minutes} = 600 \, \text{s}$
   • $n = 2$
   • $F = 96487 \, \text{C/mol}$

4. Calculate

   $m = \frac{63 \times 1.5 \times 600}{2 \times 96487}$ …

Here are the most common mistakes students make when solving this exact type of Faraday's Law electrolysis problem, along with how to avoid each.

1. Forgetting to Convert Time to Seconds

The Mistake:

Students plug in time as 10 minutes directly into $Q = I \times t$, forgetting that the standard unit for time in the formula is seconds.

How to Avoid:

Always write the conversion step explicitly:

Then use $Q = I \times t = 1.5 \times 600 = 900 \text{ C}$.

2. Using the Wrong Number of Electrons ($n$)

The Mistake:

Assuming $n = 1$ without checking the half-reaction. For copper, many students forget that $Cu^{2+}$ requires 2 electrons to become $Cu$.

How to Avoid:

Write the cathode half-reaction first:

So $n = 2$. Never guess — always write the balanced ionic equation.

3. Using the Wrong Molar Mass

The Mistake:

Using the molar mass of $CuSO_4$ (159.6 g/mol) instead of just copper ($63 \text{ g/mol}$, the value NCERT uses here).

How to Avoid:

The question asks for the mass of copper deposited, not the mass of the compound. Use $M = 63 \text{ g/mol}$ for copper.

4. Mixing Up Faraday's Constant Value

The Mistake:

Using $F = 96500$ or $F = 96487$ inconsistently, or forgetting to include it in the formula.

How to Avoid:

NCERT's Solution here uses:

Always write the full formula:

5. Arithmetic Errors in the Final Calculation

The Mistake:

Rushing the multiplication/division and getting a wrong decimal place.

How to Avoid:

Do the calculation step-by-step:

• $Q = 900 \text{ C}$
• $M = 63 \text{ g/mol}$
• $n = 2$ …