How Much Cl2 Is Consumed by a 1000 Ampere Electric Car Battery?

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Discussion Overview

The discussion revolves around calculating the amount of chlorine gas (Cl2) consumed by a 1000 Ampere electric car battery, framed within the context of electrochemistry and redox reactions. Participants explore the application of Faraday's laws and the relationship between current, charge, and moles of gas produced or consumed in a chemical reaction.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a redox reaction involving zinc and chlorine, seeking guidance on how to calculate the mass of Cl2 consumed based on the current provided.
  • Another participant suggests that the method for solving the problem is typically found in general chemistry texts, indicating a lack of immediate recall.
  • Several participants mention the importance of the Faraday constant, noting that 1 Ampere equals 1 Coulomb per second, and discuss how to relate this to the number of moles of electrons and Cl2 produced.
  • A participant calculates the consumption of chlorine gas based on the current and the stoichiometry of the reaction, arriving at a specific mass of Cl2 consumed per second.
  • Another participant provides a detailed calculation linking moles of Cl2 to Coulombs, while also noting that the calculation yields a rate of consumption rather than a total amount due to the unspecified time of use.
  • One participant expresses gratitude for the clarification regarding the relationship between amperes and coulombs, indicating a learning moment in the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the principles of electrochemistry being applied, but there are variations in the calculations and interpretations of the results. The discussion remains unresolved regarding the final amount of Cl2 consumed, as the time factor is not specified.

Contextual Notes

The calculations depend on the assumptions made about the time of current flow and the concentrations of reactants, which are not provided in the original problem statement.

chem_tr
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Hello all,

As I am preparing for my PhD proficiency exam, I tried to solve some Analytical chemistry problems. and stuck in one.

The scheme for rechargeable battery, used in automobiles working with electricity, is given below:

Zn(s)|ZnCl2 (aq) || Cl-(aq)|Cl2|C(s)

If the battery gives a current of 1.00*103 amperes, how many kilograms of Cl2 are consumed?

Well, I have attempted to solve this, but have not found enough evidence to study on and solve the problem. What I've done is basically the following:

The redox reaction might be like this:
Zn_{(s)}+Cl_2_{(aq)} \longrightarrow ZnCl_2_{(aq)}
Nernst formula may reveal something, but the final voltage as well as standard voltage (the latter can be calculated via referring to known tables, though), and we don't know anything about the concentrations. How can I use the current given? If you need, these are the half cell reactions:

Zn^{2+}+2e^- \rightleftharpoons Zn_{(s)}; E°=-0.764~Volts
Cl_2_{(aq)} + 2e^- \rightleftharpoons 2Cl^-; E°=1,395~Volts

I am in need of recommendations, thank you.
 
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Actually its quite simple...unfortunately I don't remember exactly how to do it off the top of my head. I'll get back to it later, I'm a bit busy at the moment. The method is usually introduced at the end of the electrochemistry chapter in general chemistry texts.
 
Look for Faraday constant.

Or - 1A is a Coulomb/sec, you know the current, you know the charge per second. Every two electrons give one Cl2 particle. You know the charge, you know number of the Cl2 particles consumed per second. Just convert Coulombs to Avogadro number.

That's almost the same - Faraday constant is number of Coulombs in 1 mole of electrons :)


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Last edited:
Borek said:
Look for Faraday constant.

The number that comes to mind is about 96,500 somethings ( coulombs per mole ?)
 
I know this. One faraday constant is equal to one mole of electrons - this is equal to 6.023 \times 10^{23} \times 1.6\times 10^{-19} = 96368 \frac {Coulomb}{mol}
 
Dear Borek, you said one Ampere is one coulomb per second. Then 1,000 amperes are equal to 1,000 coulombs per second. This in turn gives 6.25*1021 electrons per second, which means about 0.01 moles of electrons per second. In the redox, two electrons are used as you stated; so 0.005 moles of chlorine are consumed. This means 7.368*10-4 kilograms of chlorine gas are used.

Am I correct, pals? Thank you very much for your help, you all.
 
1 mole Cl2 == 2 moles of e- = 2*6.023*10^23*1.6*10^(-19) C = 1.927*10^5 C

1 mole Cl2 == 71 g

So 71 g Cl2 == 1.927*10^5 C

Hence 10^3 C/s == 71/192.7 = 0.368 g of Cl2 consumed per second.

I think you made a typo in the very last step (everything up to that point looks good).

Note that one can only calculate the consumption RATE, since the time of use is not given.
 
Thank you, I haven't remembered that one Ampere is equal to one Coulomb per second, so I was stuck. Again, thank you for your help.