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Difficulty: MediumIndustrial Applications of Electrolysis

In an industrial electroplating plant, a steel component is coated with silver in an electrolytic bath. If a constant electric current of 2.0 A2.0\text{ A} is passed through the bath for 48.25 minutes48.25\text{ minutes}, what is the mass of silver, in grams, deposited on the cathode? [Molar mass of Ag=108 g mol−1\text{Ag} = 108\text{ g mol}^{-1}, 1 F=96500 C mol−11\text{ F} = 96500\text{ C mol}^{-1}]

Answer: 6.48 g

Answer

The mass of silver deposited on the cathode during electroplating is 6.48 g.
According to Faraday's first law of electrolysis, charge Q=2.0 A×(48.25×60 s)=5790 CQ = 2.0\text{ A} \times (48.25 \times 60\text{ s}) = 5790\text{ C}. The number of moles of electrons passed is 5790/96500=0.06 mol5790 / 96500 = 0.06\text{ mol}. Since silver reduction (Ag++e−→Ag\text{Ag}^+ + e^- \rightarrow \text{Ag}) requires 1 mole1\text{ mole} of electrons per mole of silver, 0.06 mol0.06\text{ mol} of Ag\text{Ag} is formed. The mass of silver deposited is 0.06 mol×108 g mol−1=6.48 g0.06\text{ mol} \times 108\text{ g mol}^{-1} = 6.48\text{ g}.

Step-by-Step Solution

1
Convert time to seconds
t=48.25 min×60 s/min=2895 st = 48.25 \text{ min} \times 60 \text{ s/min} = 2895 \text{ s}
Time must be expressed in seconds to calculate electric charge in coulombs.
2
Calculate total quantity of electricity QQ
Q=I×t=2.0 A×2895 s=5790 CQ = I \times t = 2.0 \text{ A} \times 2895 \text{ s} = 5790 \text{ C}
Electric charge is the product of current and duration of electrolysis.
3
Determine moles of electrons transferred
n(e−)=5790 C96500 C mol−1=0.06 moln(e^-) = \frac{5790 \text{ C}}{96500 \text{ C mol}^{-1}} = 0.06 \text{ mol}
Faraday's constant indicates that 96500 C96500\text{ C} corresponds to 1 mole1\text{ mole} of electrons.
4
Relate electron flow to silver discharge at cathode
Ag++e−→Ag\text{Ag}^+ + e^- \rightarrow \text{Ag}, so 0.06 mol of e− yields 0.06 mol of Ag0.06 \text{ mol of } e^- \text{ yields } 0.06 \text{ mol of Ag}
Silver ion discharge requires one electron per silver atom deposited.
5
Calculate mass of silver deposited
Mass=0.06 mol×108 g mol−1=6.48 g\text{Mass} = 0.06 \text{ mol} \times 108 \text{ g mol}^{-1} = 6.48 \text{ g}
Multiplying the molar quantity by relative atomic mass yields the total mass deposited.

Key Concept

Quantitative application of Faraday's laws of electrolysis in industrial electroplating.
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Topics

Industrial Applications of Electrolysis

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