250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3 and 0.1 M

Q: Why do we divide by 96500?

Because 1 Faraday is the charge of 1 mole of electrons, which is approximately 96500 Coulombs. Dividing the total charge in Coulombs ($I \times t$) by 96500 gives us the charge in Faradays, which directly corresponds to the moles of electrons transferred.

Q: Why does gold deposit first?

Gold has a higher standard reduction potential ($1.69\text{ V}$) compared to silver ($0.80\text{ V}$). This means $\text{Au}^+$ ions have a greater tendency to gain electrons and get reduced than $\text{Ag}^+$ ions.

Q: What if we passed more current for a longer time?

If the total charge passed exceeded the amount required to deposit all the gold (0.025 F), then after all the gold has been deposited, silver would start depositing.

Q: Why did the book's solution say 0.01 moles?

That appears to be a calculation typo in the source material. Molarity $\times$ Volume in Liters ($0.1 \text{ M} \times 0.250 \text{ L}$) correctly yields 0.025 moles. However, the final conclusion remains unaffected because the passed charge (0.0093 F) is still less than the required charge.

Q: Does the 2V voltage matter in this calculation?

The 2V is the applied external voltage. As long as it is greater than the opposing cell potential (back emf) of the electrolytic cell, electrolysis will occur. For the quantitative calculation of *how much* deposits, we only need the current and time (Faraday's First Law).

Question

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M $	ext{AgNO}_3$ and 0.1 M AuCl. The solution was electrolysed at 2 V by passing a current of 1 A for 15 minutes. The metal /metals electrodeposited will be ($E^\circ_{	ext{Ag}^+/	ext{Ag}} = 0.80	ext{ V}$, $E^\circ_{	ext{Au}^+/	ext{Au}} = 1.69	ext{ V}$)

Options:

(A) only silver

(B) silver and gold in equal mass proportion

(C) only gold

(D) silver and gold in proportion to their atomic weights

Sigma Percentile · Accepted Answer

### The Goldsmith's Dilemma

Imagine you are in a goldsmith's workshop. You are handed a $250 	ext{ mL}$ beaker filled with a waste solution. But this isn't just any waste—it contains dissolved silver and gold! Specifically, it has $0.1 	ext{ M}$ of $	ext{AgNO}_3$ and $0.1 	ext{ M}$ of $	ext{AuCl}$.

Your mission is to recover these precious metals using electrolysis. You set up your electrolytic cell, apply a $2 	ext{ V}$ potential, and pass a steady current of $1 	ext{ A}$ for exactly $15 	ext{ minutes}$. The burning question is: what exactly will you find deposited on your cathode when you turn off the power?

### Calculating the Electrical Effort

Before we can determine what deposits, we need to know exactly how much electrical "effort" we put into the system. In electrochemistry, this effort is measured in Faradays, where $1 	ext{ Faraday}$ ($1 	ext{ F}$) is the charge of one mole of electrons.

We use Faraday's First Law of Electrolysis to calculate the total charge passed:

$$Q = \frac{I 	imes t}{96500}$$

Let's plug in our values. Remember, time must be in seconds!

$$Q = \frac{1 	imes (15 	imes 60)}{96500}$$

$$Q = \frac{900}{96500} \approx 0.0093 	ext{ F}$$

So, we have supplied $0.0093 	ext{ moles}$ of electrons to our cathode.

### Taking Inventory of the Ions

Next, let's figure out how much gold and silver we actually have in the beaker. We can find the number of moles by multiplying the molarity by the volume in liters.

For the silver ions ($	ext{Ag}^+$):

$$n_{	ext{Ag}^+} = 0.1 	ext{ M} 	imes 0.250 	ext{ L} = 0.025 	ext{ mol}$$

For the gold ions ($	ext{Au}^+$):

$$n_{	ext{Au}^+} = 0.1 	ext{ M} 	imes 0.250 	ext{ L} = 0.025 	ext{ mol}$$

*(Note: Some textbooks might mistakenly calculate this as $0.01 	ext{ mol}$, but the math clearly shows $0.025 	ext{ mol}$. Don't let silly mistakes derail your logic!)*

### The Race to the Cathode

When the power is turned on, both $	ext{Ag}^+$ and $	ext{Au}^+$ ions rush towards the negatively charged cathode, hungry for electrons. But they don't get to share equally. The ion with the **higher standard reduction potential** ($E^\circ$) gets priority. It's a strict hierarchy!

Let's look at the given potentials:

$$E^\circ_{	ext{Au}^+/	ext{Au}} = 1.69 	ext{ V}$$

$$E^\circ_{	ext{Ag}^+/	ext{Ag}} = 0.80 	ext{ V}$$

Since $1.69 	ext{ V} > 0.80 	ext{ V}$, gold is much more easily reduced than silver. Therefore, **gold will be preferentially discharged** at the cathode.

### The Final Verdict

Now for the final check. We know gold will deposit first, but will we have enough electrons left over for silver to start depositing?

The reduction reaction for gold is:

$$	ext{Au}^+(aq) + e^- ightarrow 	ext{Au}(s)$$

This tells us that $1 	ext{ mole}$ of $	ext{Au}^+$ requires $1 	ext{ mole}$ of electrons ($1 	ext{ F}$) to fully deposit. Since we have $0.025 	ext{ moles}$ of $	ext{Au}^+$, we would need exactly $0.025 	ext{ F}$ of charge to recover all the gold.

However, we calculated earlier that we only supplied $0.0093 	ext{ F}$ of charge.

$$0.0093 	ext{ F} < 0.025 	ext{ F}$$

We don't even have enough electrons to deposit all the gold! The battery will effectively "run out" of the specified time before the gold is fully recovered. Consequently, the silver ions will just have to keep swimming. **Only gold** will be electrodeposited.

Select Answer:

Visualized Solution

\text{Electrolytic Setup}

\text{Total Charge Passed}

\text{Calculating Charge}

\text{Moles of Ions}

\text{Preferential Discharge}

\text{Cathode Reaction}

\text{Conclusion}

The Sigma Insight: Electrolytic Conduction

The Goldsmith's Dilemma

Calculating the Electrical Effort

Taking Inventory of the Ions

The Race to the Cathode

The Final Verdict

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