Q.Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.

The key idea is Faraday’s laws of electrolysis: the mass of a substance deposited at an electrode is proportional to its equivalent weight ($E = M/n$, where $M$ is molar mass and $n$ is the number of electrons transferred per ion).

Reasoning:

1. Same current ($1\ \text{A}$) and same time ($10\ \text{min} = 600\ \text{s}$) means the same quantity of charge passes through both cells: $Q = I t = 1 \times 600 = 600\ \text{C}$. …

Why Faraday’s laws answer this directly

When you pass the same current for the same time through two different electrolytic cells connected in series (or run separately with identical current and time), the quantity of charge flowing through each is identical:

Faraday’s first law says: the mass of a substance deposited at an electrode is proportional to the charge passed. But the constant of proportionality depends on the substance — specifically on its electrochemical equivalent $Z$, which is:

where $M$ is molar mass, $n$ is the number of electrons transferred per ion (the valency factor), and $F = 96485\ \text{C mol}^{-1}$ is Faraday’s constant.

So the mass deposited is:

Since $Q$ and $F$ are the same for both cells, the ratio of masses depends only on $\frac{M}{n}$ — the equivalent weight.

Step-by-step reasoning

1. Identify the cathode reactions

   In aqueous solution:

   • Copper sulphate: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ Here $n = 2$, $M_{\text{Cu}} = 63.5\ \text{g mol}^{-1}$.
   • Silver nitrate: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ Here $n = 1$, $M_{\text{Ag}} = 108\ \text{g mol}^{-1}$.

2. Compute the equivalent weights

   • Equivalent weight of Cu = $\frac{63.5}{2} = 31.75\ \text{g eq}^{-1}$
   • Equivalent weight of Ag = $\frac{108}{1} = 108\ \text{g eq}^{-1}$
   Watch out

   A common mistake is to compare molar masses directly (63.5 vs 108) and conclude copper deposits less. That’s correct here, but only because silver’s n-factor is smaller. If both had the same n-factor, the comparison would be different.

3. Calculate the masses deposited

   Using $m = \frac{M}{nF} \times Q$:

   • For copper:

• For silver: …

Faraday's Laws of Electrolysis — Solution Method

Method: Faraday's Second Law of Electrolysis (Law of Equivalent Masses)

Step 1: State the Law

Faraday's Second Law states:

When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are directly proportional to their chemical equivalents (equivalent masses).

Mathematically:

where:

• $m$ = mass deposited
• $E$ = equivalent mass of the substance

Step 2: Identify the Given Data

• Current ($I$) = 1 A (same for both cells)
• Time ($t$) = 10 minutes = $10 \times 60 = 600$ seconds (same for both cells)
• Electrolytes: Aqueous $\text{CuSO}_4$ and aqueous $\text{AgNO}_3$

Since current and time are identical, the quantity of electricity ($Q = I \times t$) passed is the same for both cells.

Step 3: Find Equivalent Masses

For copper in $\text{CuSO}_4$:

• Copper ion: $\text{Cu}^{2+}$ (valency = 2)
• Atomic mass of Cu = 63.5 g/mol
• Equivalent mass $E_{\text{Cu}} = \frac{\text{Atomic mass}}{\text{Valency}} = \frac{63.5}{2} = 31.75$ g/eq

For silver in $\text{AgNO}_3$:

• Silver ion: $\text{Ag}^{+}$ (valency = 1)
• Atomic mass of Ag = 108 g/mol
• Equivalent mass $E_{\text{Ag}} = \frac{108}{1} = 108$ g/eq

Step 4: Apply the Law …

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 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 $Q = I \times t = 1 \times 600 = 600 \text{ C}$ (the current given in the question is 1 ampere, not 1.5 A — always re-check the given data before plugging in numbers).

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

The Mistake:

Assuming $n = 1$ for both metals, or guessing 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-reactions first:

Never guess — always write the balanced ionic equation for each metal separately.

3. Using the Wrong Molar Mass

The Mistake:

Using the molar mass of the compound (e.g. $CuSO_4$, 159.6 g/mol) instead of just the metal (63.5 g/mol for Cu, 108 g/mol for Ag).

How to Avoid:

The question asks for the mass of metal deposited, not the mass of the compound. Use $M_{Cu} = 63.5$ g/mol and $M_{Ag} = 108$ g/mol.

4. Mixing Up Faraday's Constant Value

The Mistake:

Using $F = 96500$ or $F = 96485$ inconsistently, or forgetting to include it in the formula.

How to Avoid:

Memorise:

Always write the full formula:

5. Arithmetic Errors — Especially Using the Wrong Given Current

The Mistake:

Rushing through the calculation and substituting the wrong current value (e.g. using 1.5 A when the question states 1 A), which throws off the entire final answer.

How to Avoid:

Do the calculation step-by-step, always re-checking the given data first:

• $I = 1\ \text{A}$ (as stated in the question)
• $t = 600\ \text{s}$
• $Q = I \times t = 600\ \text{C}$

For copper ($M=63.5$, $n=2$, $F=96500$): …