Numerical/Formula

1. TITLE

How to solve Faraday's First Law Mass Deposited numericals in NEET — Step-by-step

2. PROBLEM TYPE INTRO

This category of numerical applies Faraday's First Law of Electrolysis to calculate the amount of a substance deposited at an electrode. You will need the formula $Q = It$ alongside standard reaction stoichiometry. The typical mistake students make is forgetting to convert time into seconds or misidentifying the metal's valency.

3. MCQ

A solution of $\text{CuSO}_4$ is electrolysed for $10\text{ minutes}$ with a current of $1.5\text{ amperes}$. What is the mass of copper deposited at the cathode? (Given: Atomic mass of Cu = $63\text{ g/mol}$, $F = 96487\text{ C mol}^{-1}$)

A) $0.294\text{ g}$
B) $0.588\text{ g}$ (Incorrect: Used wrong valency of n=1 instead of n=2)
C) $0.0049\text{ g}$ (Incorrect: Forgot to convert minutes to seconds)
D) $0.890\text{ g}$ (Incorrect: Arbitrary calculation error)

4. STEP-BY-STEP SOLUTION

SOLUTION

Given:
* Current ($I$) = $1.5\text{ A}$
* Time ($t$) = $10\text{ minutes}$
* Atomic mass of Cu = $63\text{ g/mol}$
* Faraday constant ($F$) = $96487\text{ C mol}^{-1}$

To Find:
* Mass of copper deposited at the cathode ($w$)

Formula:
* $Q = It$
Use this to calculate the total quantity of electricity (charge) passed through the electrolyte.
* $w = \frac{\text{Atomic Mass} \times Q}{n \times F}$
Use this to relate the total charge to the mass of the element deposited based on its stoichiometry.

Step 1: Calculate total charge passed ($Q$) in coulombs
Convert time to seconds: $t = 10 \times 60 = 600\text{ s}$.
$Q = 1.5\text{ A} \times 600\text{ s} = 900\text{ C}$.

Step 2: Determine the valency ($n$) from the electrode reaction
The reduction reaction at the cathode for copper sulphate is $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$.
This means $n = 2$ moles of electrons are required to deposit 1 mole of copper.

Step 3: Calculate the mass of copper deposited
We know $2F$ ($2 \times 96487\text{ C}$) deposits $1\text{ mol}$ ($63\text{ g}$) of Cu.
For $900\text{ C}$, the mass deposited is:
$w = \frac{63\text{ g} \times 900\text{ C}}{2 \times 96487\text{ C mol}^{-1}}$
$w = 0.2938\text{ g}$

Answer: $0.294\text{ g}$

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1: Forgetting to convert time from minutes or hours into seconds before applying $Q = It$.
  • ❌ Mistake 2: Using the wrong valency (n-factor). Students often forget that $\text{Cu}^{2+}$ requires $2$ moles of electrons, blindly calculating using $n=1$.
  • ❌ Mistake 3: Using an incorrect value for the Faraday constant. Always stick to the standard $96487\text{ C mol}^{-1}$ or the approximate $96500\text{ C mol}^{-1}$ if options are spread out.

6. QUICK FORMULA REFERENCE

Formulas Used:
* $Q = I \times t$ — Use when finding total electrical charge passed (always ensure $I$ is in Amperes and $t$ is in seconds).
* $w = \frac{M \times Q}{n \times F}$ — Use when converting the calculated charge ($Q$) into the mass ($w$) of metal deposited at the electrode.

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: Calculating the mass of silver (Ag) deposited from an $\text{AgNO}_3$ solution using a given current and time (remember, silver undergoes the reaction $\text{Ag}^+ + e^- \rightarrow \text{Ag}$, so its valency is $n=1$!)".

8. NCERT REFERENCE

Based on: NCERT Class 12, Chapter 2 (Electrochemistry), Example 2.10.

9. CTA

More solved numericals → Redox Reactions & Electrochemistry Complete Guide
Practice problems → Mock exams