Calculate the mass of Ag deposited at cathode when a current of 1 ampere was passed through a solution of AgNO3 for 15 min. How many electrons flow through a metallic wire if a cur... Calculate the mass of Ag deposited at cathode when a current of 1 ampere was passed through a solution of AgNO3 for 15 min. How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours?
Understand the Problem
The question involves calculating the mass of silver deposited at the cathode during electrolysis using specific parameters such as current, time, and the molar mass of silver. It also touches on the flow of electrons and the application of Faraday's laws of electrolysis.
Answer
The mass of Ag deposited is approximately $0.2 \, \text{g}$ and the number of electrons is approximately $2.25 \times 10^{22}$.
Answer for screen readers
The mass of Ag deposited is approximately 0.2 g, and the total number of electrons that flowed through the wire is approximately $2.25 \times 10^{22}$ electrons.
Steps to Solve
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Calculate the mass of Ag deposited at the cathode
Use Faraday's first law of electrolysis, which states: $$ \text{Mass} = \frac{\text{Molar Mass} \times \text{Current} \times \text{Time}}{nF} $$
Given:
- Current ($I$) = 1 A
- Time ($t$) = 15 min = 15 × 60 s = 900 s
- Molar mass of Ag = 108 g/mol
- Number of electrons transferred ($n$) for Ag = 1 (from the reaction $Ag^+ + e^- \rightarrow Ag$)
- Faraday's constant ($F$) = 96500 C/mol
Plug in the values: $$ \text{Mass} = \frac{108 , \text{g/mol} \times 1 , \text{A} \times 900 , \text{s}}{1 \times 96500 , \text{C/mol}} $$
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Calculate the number of electrons flowing through the wire
Now, for the current of 0.5 A passed for 2 hours:
- Convert time to seconds: $t = 2 , \text{hours} = 2 \times 3600 , \text{s} = 7200 , \text{s}$
- Total charge ($Q$) can be calculated with: $$ Q = I \times t $$
Using: $$ Q = 0.5 , \text{A} \times 7200 , \text{s} = 3600 , \text{C} $$
To find the number of electrons, use: $$ \text{Number of electrons} = \frac{Q}{e} $$
Where $e = 1.6 \times 10^{-19} , \text{C}$ (charge of one electron). Therefore, $$ \text{Number of electrons} = \frac{3600 , \text{C}}{1.6 \times 10^{-19} , \text{C/electron}} $$
The mass of Ag deposited is approximately 0.2 g, and the total number of electrons that flowed through the wire is approximately $2.25 \times 10^{22}$ electrons.
More Information
The calculation uses basic principles of electrolysis and Faraday’s laws, a foundational concept in electrochemistry. The law relates the amount of substance deposited during electrolysis to the quantity of electric charge passed through the substance.
Tips
- Incorrect unit conversion: Always ensure time is in seconds when calculating charge.
- Forgetting Faraday's constant: It's essential for the mass calculation in electrolysis.
- Not adjusting for number of electrons: Ensure that $n$ is correct for the ion being reduced or oxidized.
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