Q.A solution of $Ni(NO_3)_2$ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

The key idea is Faraday’s laws of electrolysis: the mass of a substance deposited is proportional to the total charge passed and its equivalent weight.

Step 1 — Total charge passed

Current $I = 5\ \text{A}$, time $t = 20\ \text{min} = 20 \times 60 = 1200\ \text{s}$.

Charge $Q = I \times t = 5 \times 1200 = 6000\ \text{C}$.

Step 2 — Moles of electrons

$1\ \text{F} = 96500\ \text{C}$ carries 1 mole of electrons.

Moles of electrons $= \frac{6000}{96500} \approx 0.06218\ \text{mol}$.

Step 3 — Moles and mass of Ni

Nickel deposition: $\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$ …

This is a straightforward application of Faraday’s laws of electrolysis. The key idea: the amount of substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. For nickel, which is divalent ($Ni^{2+}$), each mole of Ni requires 2 moles of electrons.

Let’s work through it step by step.

1. Find the total charge passed.

   Current $I = 5$ A, time $t = 20$ minutes $= 20 \times 60 = 1200$ s.

   Charge $Q = I \times t = 5 \times 1200 = 6000$ coulombs.

2. Relate charge to moles of electrons.

   Faraday’s constant $F = 96500$ C/mol e$^-$ (the charge on one mole of electrons).

   Moles of electrons $= \dfrac{Q}{F} = \dfrac{6000}{96500} \approx 0.06218$ mol.

3. Convert moles of electrons to moles of Ni.

   The half-reaction at the cathode:

So 2 moles of electrons deposit 1 mole of Ni.

Moles of Ni $= \dfrac{\text{moles of electrons}}{2} = \dfrac{0.06218}{2} = 0.03109$ mol.

4. Convert moles of Ni to mass. Atomic mass of Ni $= 58.7$ g/mol (standard value used in such problems). Mass $= \text{moles} \times \text{molar mass} = 0.03109 \times 58.7 \approx 1.825$ g. …

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

• $m$ = mass deposited (in grams)
• $Q$ = total charge passed (in coulombs)

Steps to solve

Step 1: Calculate total charge ($Q$)

Current $I = 5 \, \text{A}$, time $t = 20 \, \text{minutes} = 20 \times 60 = 1200 \, \text{s}$

Step 2: Write the cathode reaction

Nickel ions are reduced at the cathode:

This tells us:

• 2 moles of electrons are needed to deposit 1 mole of Ni
• Charge of 1 mole of electrons = $1 \, F = 96500 \, \text{C}$

So, charge required to deposit 1 mole of Ni = $2 \times 96500 = 193000 \, \text{C}$

Step 3: Apply Faraday’s First Law

where

• $M$ = molar mass of Ni = $58.7 \, \text{g/mol}$ …

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

How to Avoid:

Always write the conversion step explicitly:

Then use $Q = I \times t = 5 \times 1200 = 6000 \text{ C}$.

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

The Mistake:

Assuming $n = 1$ without checking the half-reaction. Nickel, like copper, is divalent and needs 2 electrons to be deposited.

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 $Ni(NO_3)_2$ instead of just nickel metal.

How to Avoid:

The question asks for mass of nickel deposited, not the mass of the salt. Use $M = 58.7 \text{ g/mol}$ for Ni.

4. Mixing Up Faraday’s Constant Value

The Mistake:

Using $F = 96500$ and $F = 96485$ inconsistently within the same calculation.

How to Avoid:

Pick one value and use it throughout:

5. Arithmetic Errors / Premature Rounding

The Mistake:

Rounding moles of electrons or moles of Ni too early, shifting the final mass.

How to Avoid:

Do the calculation step-by-step, keeping several significant figures until the end: …