How much gold is deposited in 6 hours of electrolysis at 0.540 A?

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Homework Help Overview

The discussion revolves around the electrolysis of a gold ion solution (Au3+) to determine the amount of gold deposited over a period of 6 hours at a constant current of 0.540 A. Participants are exploring the application of Faraday's law in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of moles of electrons using the formula It/F and question the next steps after this calculation. There is mention of using Faraday's law and the half-reaction for gold deposition to relate charge to the amount of gold deposited.

Discussion Status

Some participants have offered guidance on using specific formulas and have shared their calculations, while others are still clarifying their understanding of the process. Multiple interpretations of the problem and methods are being explored without a clear consensus on the final approach.

Contextual Notes

There is a mention of the formulation being vague, and one participant notes the possibility of leaving the final answer in kilograms, indicating some uncertainty about the expected units for the result.

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In the electrolysis of Au3+ (aq) solution, gold is deposited. How much gold is deposited in 6 hours by a constant current of 0.540 A?

I know you need to find the moles of e- by using It/F. But I'm not sure where to go from there.

Any help would be great.

Thanks.
 
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Use Faraday's law,compute the mass of Au by making those multiplications/divisions and then compare to the fninal result.

The formulation's kinda vague,so you can leave the final answer in Kg...

Daniel.
 
Consider the half-reaction

Au3+(aq) + 3e- -------------> Au(s)

Now 1 mol of Au3+ requires 3 Faradays (3*96500 C) to deposit 1 mol of Au

Therefore, if the amount of charge passed is 6*3600*.540 C, how many moles of Au will be deposited?
 
Okie dokie, thanks.
 
I just used n = It/zF (where z= amount of "excess" electrons). Then multiplied the answer by the molar mass of gold. That gave me 7.94g.
 

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