What mass of silver is deposited during electrolysis when a current of 0.8 A flows for 25 minutes?

Analysis

To solve this problem, we use Faraday’s First Law of Electrolysis, which states that the mass ($m$) of a substance deposited during electrolysis is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte.

Concept

The formula is:

Where:

• $m$ = mass deposited (in grams)
• $I$ = current (in Amperes) = $0.8\text{ A}$
• $t$ = time (in seconds) = $25\text{ minutes} \times 60 = 1500\text{ seconds}$
• $M$ = Molar mass of Silver (Ag) $\approx 108\text{ g/mol}$
• $n$ = Valency of Silver = $1$ (since $\text{Ag}^+ + e^- \rightarrow \text{Ag}$)
• $F$ = Faraday’s constant $\approx 96,500\text{ C/mol}$

Step-by-Step Solution

1. Calculate the quantity of electricity ($Q$): $Q = I \times t$ $Q = 0.8\text{ A} \times 1500\text{ s} = 1200\text{ Coulombs}$

2. Apply the formula: $m = \frac{Q \times M}{n \times F}$ $m = \frac{1200 \times 108}{1 \times 96500}$

3. Perform the calculation: $m = \frac{129600}{96500}$ $m \approx 1.343\text{ g}$

Rounding to two decimal places, we get 1.34 g.

Conclusion

Comparing our result to the options provided, the calculated mass matches option D.

Exam Tip: Always remember to convert time from minutes to seconds before plugging it into the formula. Many students lose marks by forgetting the $\times 60$ conversion! Also, keep the Faraday constant ($96,500\text{ C}$) in your memory as it is a standard value in JAMB Physics and Chemistry.