Q.Three electrolytic cells A, B, C containing solutions of $ZnSO_4$, $AgNO_3$ and $CuSO_4$, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Concept: Faraday’s laws of electrolysis — the mass of a substance deposited is proportional to the charge passed and its equivalent weight.

Reasoning:

1. For silver (Ag, atomic mass 108, valency 1), equivalent weight = 108 g/eq.

   Charge required for 1.45 g Ag:

   $Q = \frac{1.45 \times 96500}{108} \approx 1296\ \text{C}$.

2. Current $I = 1.5\ \text{A}$, so time $t = \frac{Q}{I} = \frac{1296}{1.5} = 864\ \text{s}$.

3. For copper (Cu, atomic mass 63.5, valency 2), equivalent weight = 31.75 g/eq.

   Mass deposited: $m_{\text{Cu}} = \frac{31.75 \times 1296}{96500} \approx 0.426\ \text{g}$. …

When cells are connected in series, the same current flows through each for the same time. That means the quantity of charge (in coulombs) that passes through one cell is identical to that passing through every other cell. This is the key to solving series electrolysis problems: you only need to find the charge from one cell’s data, then apply it to the others.

Faraday’s first law says: mass deposited is directly proportional to charge passed. Mathematically, $m = Z \cdot Q$, where $Z$ is the electrochemical equivalent (mass deposited per coulomb). But it’s often easier to use the molar form: $m = \frac{Q \cdot M}{n \cdot F}$, where $M$ is molar mass, $n$ is the number of electrons involved per ion, and $F$ is Faraday’s constant (96485 C/mol). The quantity $\frac{M}{n}$ is the equivalent mass.

Let’s work through it step by step.

1. Find the charge from the silver deposited. Silver is deposited from $AgNO_3$: $Ag^+ + e^- \rightarrow Ag$. So $n = 1$. Molar mass of Ag = 107.87 g/mol (we’ll use 108 g/mol for simplicity, as is common in exams). Mass deposited $m_{Ag} = 1.45$ g. Using $m = \frac{Q \cdot M}{n \cdot F}$, we get

Calculate: $1.45 \times 96485 = 139903.25$, divide by 108 gives $Q \approx 1295.4$ C.

2. Find the time of current flow. Current $I = 1.5$ A. Charge $Q = I \cdot t$, so

Convert to minutes: $863.6 / 60 \approx 14.39$ minutes.

So the current flowed for about 14.4 minutes.

3. Find the mass of copper deposited. In cell C, $CuSO_4$ gives $Cu^{2+} + 2e^- \rightarrow Cu$, so $n = 2$. Molar mass of Cu = 63.55 g/mol (often taken as 63.5 in problems). The same charge $Q = 1295.4$ C passes through this cell.

Numerator: $1295.4 \times 63.55 \approx 82322.67$; denominator: $2 \times 96485 = 192970$.

So $m_{Cu} \approx 0.4266$ g. …

Faraday’s Laws of Electrolysis — Series Cell Problem

Method: Faraday’s Second Law (Law of Equivalent Masses)

When cells are connected in series, the same quantity of electricity passes through each cell. Therefore, the masses of different substances deposited are directly proportional to their chemical equivalents (equivalent weights).

Step-by-step solution

Step 1: Find the time of current flow using cell B (silver)

Given:

• Current, $I = 1.5 \, \text{A}$
• Mass of silver deposited, $m_{\text{Ag}} = 1.45 \, \text{g}$
• Atomic mass of Ag = $108 \, \text{g/mol}$
• Valency of Ag = $+1$ (since $\text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^-$)

Equivalent weight of Ag = $\frac{108}{1} = 108 \, \text{g/eq}$

Using Faraday’s First Law:

where $E$ = equivalent weight, $t$ = time in seconds.

Rearranging:

Substitute:

Answer: $\boxed{864 \, \text{s}}$ (or 14 minutes 24 seconds)

Step 2: Find mass of copper deposited (cell C)

• Atomic mass of Cu = $63.5 \, \text{g/mol}$
• Valency of Cu in $\text{CuSO}_4$ = $+2$
• Equivalent weight of Cu = $\frac{63.5}{2} = 31.75 \, \text{g/eq}$

Since same charge passes through all cells, using Faraday’s Second Law:

Answer: $\boxed{0.426 \, \text{g}}$

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Here are the common mistakes students make on this classic Faraday’s Laws of Electrolysis problem, and how to avoid each.

1. Mistake: Forgetting the “Series” Connection Implies Equal Charge

The error: Students calculate time using only the silver data, but then use different charges for copper and zinc.

Why it’s wrong: In a series circuit, the same current flows for the same time through all cells. That means the quantity of charge ($Q = I \times t$) passed is identical for all three cells.

How to avoid:

• Always write: $Q$ is the same for A, B, and C.
• Calculate $Q$ from the silver data first, then use that same $Q$ for copper and zinc.

2. Mistake: Using the Wrong Molar Mass or Valency

The error: Using atomic mass of Ag as 108 g/mol (correct) but then using Cu = 63.5 g/mol and Zn = 65.4 g/mol without checking the oxidation state in the electrolysis.

Why it’s wrong: Faraday’s law uses equivalent mass = $\frac{\text{Molar mass}}{n}$, where $n$ is the number of electrons transferred per ion.

How to avoid:

• Always write the half-reaction first.
• Use $n = 1$ for Ag, $n = 2$ for Cu and Zn.
• Then apply:

where $F = 96500\ \text{C/mol}$.

3. Mistake: Incorrect Unit Conversion for Time

The error: After finding $Q = I \times t$, students forget to convert seconds to minutes/hours or vice versa.

Example: $Q = 1297.5\ \text{C}$, $I = 1.5\ \text{A}$ → $t = \frac{1297.5}{1.5} = 865\ \text{s}$.

Some write 865 minutes or 865 hours.

How to avoid:

• Use: $t = \frac{Q}{I}$ with $Q$ in coulombs, $I$ in amperes → $t$ in seconds.
• Convert only if the question asks for minutes/hours.
• Check reasonableness: 1.5 A for ~14 minutes is plausible; 1.5 A for 14 hours is not.

4. Mistake: Using the Mass of Silver to Directly Find Mass of Copper/Zinc by Simple Proportion

The error: Thinking “if 1.45 g Ag deposits, then same mass of Cu deposits” or using a ratio of molar masses only.

Why it’s wrong: Faraday’s law depends on equivalents, not just mass.

• 1 mole of Ag (107.87 g) requires 1 mole of electrons.
• 1 mole of Cu (63.55 g) requires 2 moles of electrons. So for the same charge, mass of Cu deposited is less than half the mass of Ag.

How to avoid:

• Always use the formula:

or compute each mass separately from $Q$.

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