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Chapter 19: Problem 107

A spoon was silver-plated electrolytically in a \(\mathrm{AgNO}_{3}\) solution. (a) Sketch a diagram for the process. (b) If \(0.884 \mathrm{~g}\) of \(\mathrm{Ag}\) was deposited on the spoon at a constant current of \(18.5 \mathrm{~mA},\) how long (in minutes) did the electrolysis take?

Short Answer

Expert verified
The time taken for electrolysis to deposit \(0.884g\) of silver on the spoon is 711 minutes.

Step by step solution

01

Understand Electrolysis and Sketch the Process

Electrolysis is a process in which electrical energy is used to cause a non-spontaneous chemical reaction. In this case, a spoon is coated with silver using silver nitrate solution (\(\mathrm{AgNO}_{3}\)). Silver is the cation in the silver nitrate solution that accepts electrons at the cathode to become silver metal and get deposited on the spoon. The spoon acts as a cathode in this process where reduction of silver ions (\(Ag^+\)) happens. The anode gets oxidized releasing electrons that flow through the external circuit. An inert electrode (like a graphite rod) can be used as an anode.
02

Understand Faraday's law of Electrolysis

To answer the second part, we should understand Faraday's law of electrolysis. It states that the amount of substance deposited on an electrode after electrolysis is proportional to the amount of electricity passed. This is commonly given by the formula: \( Q = It \), where: Q is the amount of charge, I is the current, and t is the time.
03

Calculate the Number of Moles of Silver Deposited

We also need to determine the amount of silver deposited on the spoon in terms of moles because 1 mole electron is required to deposit 1 mole of \(Ag\). The molar mass of silver (\(Ag\)) is approximately 107.87 g/mol. Therefore, \(0.884g\) of \(Ag\) is equivalent to \(\frac {0.884g}{107.87 g/mol} = 0.00819 mol\)
04

Determine the Charge Transferred

Since 1 mole of electrons corresponds to 1 Faraday of charge (\(F = 96485 C/mol\)) and we need 1 mole of electrons to deposit 1 mole of \(Ag\), the amount of charge required to deposit \(0.00819 mol\) of \(Ag\) is given by: \(Q = 0.00819 mol * 96485 C/mol = 790 C\)
05

Calculate the Time Taken for Electrolysis

Use the formula \(Q = It\) to find out the time. Given the current I = 18.5 mA = 0.0185 A, we can rearrange the formula to solve for time: \(t = \frac{Q}{I} = \frac{790 C}{0.0185 A} = 42649 s\). To convert seconds into minutes divide by 60 which gives \(t = \frac{42649 s}{60} = 711 min\).

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