When we calculate amperes over a certain period of time, is that the average amperes? Because we learned that instantaneous E changes as reaction gets closer to equilibrium, and since the concentration of reactants decrease over time, I'd imagine that the reaction gets slower, impacting amperes. If that's the case, then is it average?

If this were a voltaic cell, then yes the current would decrease as the reaction progressed. However, here we have an electrolytic cell where an external power source is supplying a constant voltage (and therefore a constant current). So the current is constant here because the voltage isn’t originating from the reaction itself, but a battery. Hope that helps.

Ok, so I am putting this question here because it is generally about Faraday's constant and all that. Let's say your calculating your change in free energy(delta G). You have the equation delta G=-nFE where n is the moles of electrons, F is the faraday constant, and E is the cell potential(positive for galvanic cells). Does that negative sign apply to the whole equation? And if so, why does the faraday constant never have negative sign if the negative sign is applied to the whole equation? Thanks so much for whoever is gonna help me in advance!

Before I answer the question I just want to point out that the equation you listed only applies for standard free energy change, ΔG°. Which means that the cell potential is also a standard cell potential, E°. The Nernst equation is how we would relate equilibrium to cell potential and is analogous to the nonstandard Gibbs free energy equation. Using the equation ΔG° = -nFE°, we can think of the negative sign as multiplying the other three terms by -1. So we could also write it as: ΔG° = -1*nFE°. The negative sign only applies to the -1 and not the other three terms. The only reason it’s there is that it makes positive cell potentials correlate to negative changes in free energy since both represent spontaneous reactions. Mathematical constants, like Faraday’s constant, are always listed as positive numbers. This eliminates any confusion when using constants as to whether we have to use a positive number or a negative number in formulae. Hope that helps.

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Class 12 Chemistry (India)

Course: Class 12 Chemistry (India) > Unit 2

Lesson 5: Quantitative aspect of electrolysis - Faraday's Laws

Quantitative electrolysis

Name: Quantitative electrolysis
Uploaded: 2022-01-11T22:17:00.362877679Z
Description: Given the amount of electrical charge that passes through an electrolytic cell, we can calculate the quantity of substances consumed or produced during electrolysis (or vice versa). The total charge is related to the magnitude of the current and the time it runs by the equation I = Q/t, where I is the current in amperes, Q is the charge in coulombs, and t is the time in seconds.

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Given the amount of electrical charge that passes through an electrolytic cell, we can calculate the quantity of substances consumed or produced during electrolysis (or vice versa). The total charge is related to the magnitude of the current and the time it runs by the equation I = Q/t, where I is the current in amperes, Q is the charge in coulombs, and t is the time in seconds. Created by Jay.

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Vinavs
Posted 8 months ago. Direct link to Vinavs's post “When we calculate amperes...”
When we calculate amperes over a certain period of time, is that the average amperes? Because we learned that instantaneous E changes as reaction gets closer to equilibrium, and since the concentration of reactants decrease over time, I'd imagine that the reaction gets slower, impacting amperes. If that's the case, then is it average?
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(2 votes)
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- Richard
  Posted 8 months ago. Direct link to Richard's post “If this were a voltaic ce...”
  If this were a voltaic cell, then yes the current would decrease as the reaction progressed. However, here we have an electrolytic cell where an external power source is supplying a constant voltage (and therefore a constant current). So the current is constant here because the voltage isn’t originating from the reaction itself, but a battery.
  
  Hope that helps.
  Comment on Richard's post “If this were a voltaic ce...”
  (3 votes)
Forever Learner
Posted 8 months ago. Direct link to Forever Learner's post “Ok, so I am putting this ...”
Ok, so I am putting this question here because it is generally about Faraday's constant and all that. Let's say your calculating your change in free energy(delta G). You have the equation delta G=-nFE where n is the moles of electrons, F is the faraday constant, and E is the cell potential(positive for galvanic cells). Does that negative sign apply to the whole equation? And if so, why does the faraday constant never have negative sign if the negative sign is applied to the whole equation? Thanks so much for whoever is gonna help me in advance!
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Richard
  Posted 8 months ago. Direct link to Richard's post “Before I answer the quest...”
  Before I answer the question I just want to point out that the equation you listed only applies for standard free energy change, ΔG°. Which means that the cell potential is also a standard cell potential, E°. The Nernst equation is how we would relate equilibrium to cell potential and is analogous to the nonstandard Gibbs free energy equation.
  
  Using the equation ΔG° = -nFE°, we can think of the negative sign as multiplying the other three terms by -1. So we could also write it as: ΔG° = -1*nFE°. The negative sign only applies to the -1 and not the other three terms. The only reason it’s there is that it makes positive cell potentials correlate to negative changes in free energy since both represent spontaneous reactions.
  
  Mathematical constants, like Faraday’s constant, are always listed as positive numbers. This eliminates any confusion when using constants as to whether we have to use a positive number or a negative number in formulae.
  
  Hope that helps.
  Comment on Richard's post “Before I answer the quest...”
  (2 votes)

Video transcript

- [Instructor] We already know that in an electrolytic cell, current or movement of electrons is used to drive a redox reaction. And if we look at a generic reduction half-reaction, the stoichiometry of the half reaction shows how many electrons are needed to reduce a generic metal ion. For example, if we say that our generic metal ion is M2+, it takes two electrons to turn M2+ into the solid metal. So one mole of M2+ ions requires two moles of electrons to form one mole of the metal. So we could use this mole ratio of one mole of metal to two moles of electrons to figure out, say, how much of a metal is produced in an electrolytic reaction. So this is an example of a quantitative electrolysis problem. Moles of electrons are related to the amount of charge that passes through the electrolytic cell because one mole of electrons carries 96,485 coulombs. So we could set this up as a constant called Faraday's constant. So Faraday's constant is symbolized by F, and it's equal to 96,485 coulombs per mole of electrons. And charge is related to current and time by this equation. So I is equal to Q over t; where I is the current; Q, capital Q is the charge; and t is the time. Sometimes, you see a lower case Q for charge. Charge is measured in coulombs, and time is in seconds. And one coulomb per second is equal to one ampere, so current is measured in amperes. Next, let's look at a conversion chart that shows how to approach quantitative electrolysis problems. And let's start over here with this box on the far left, which says that charge is equal to current times time. This comes from our equation for current, I is equal to Q over t. So solving for charge, solving for Q, Q is equal to I times t or charge is equal to current times time. Charge is related to moles of electrons by Faraday's constant. So we can convert back and forth between the two using Faraday's constant. And moles of electrons are related to the moles of the substance that we're interested in the problem by the balanced half-reaction. So the balanced half-reaction allows us to convert back and forth between moles of electrons and moles of substance. And from stoichiometry, we know that the moles of a substance is related to the grams of the substance by the molar mass. So the molar mass allows us to convert back and forth between these quantities. So it helps to look at a conversion chart when doing these sorts of problems. However, if you don't remember the chart, you can always use dimensional analysis and units to figure out the answer to the question. Next, let's use our conversion chart to figure out a quantitative electrolysis problem. Let's say we have an aqueous solution of silver nitrate and a constant current of 2.40 amperes is applied for 1,225 seconds. And our goal is to figure out the mass of silver that has formed. As a quick reminder, an aqueous solution of silver nitrate contains Ag+ ions and nitrate anions in solution. So the applied current is used to reduce the Ag+ ions to form solid silver. And since our goal is to figure out the mass of the solid silver that's formed, we're trying to find grams of substance. And in the problem we're given the current and the time, so we're starting over here in the first box with the current and the time. So let's think about the steps that we need to solve this problem. The first step is to find the charge, which we can get from the current and the time. The second step is to convert charge into moles of electrons. From moles of electrons, we can convert that into the moles of silver in this case. And in the fourth step for this problem, we can convert moles of silver into grams of silver. So step one is to find the charge. And Q, the charge, is equal to the current times the time. So the current is equal to 2.40 amperes. And remember, one ampere is equal to one coulomb per second. So we plug in 2.40 coulombs per second. And the time is 1,225 seconds, so we plug that in, seconds cancels out, and we get that Q is equal to 2,940 coulombs. So that's the charge that's transferred in that time period. Now that we have the charge, in the second step, our goal is to find the moles of electrons. And charge is related to moles of electrons by Faraday's constant. So dividing the charge by Faraday's constant causes coulombs to cancel out, and we get 0.0305 moles of electrons. And now that we have moles of electrons, in step three, we can find the moles of silver. Looking at the balanced equation for our half reaction showing the reduction of silver ions, one mole of silver ions requires one mole of electrons to form one mole of solid silver. So our mole ratio of silver to moles of electrons is one-to-one, which gives us this conversion factor of one mole of silver per one mole of electrons. So if we multiply our moles of electrons by our conversion factor, moles of electrons cancels out, and that gives us 0.0305 moles of silver. In this case, our mole ratio was one-to-one, but if we were reducing a different metal ion, our mole ratio might not be one-to-one. So we have to be extra careful on the step to make sure we get the right mole ratio. And since we have moles of silver, in our fourth and final step, we can convert moles of silver into grams of silver. So multiplying moles of silver by the molar mass of silver, which is 107.87 grams per mole, moles cancel and this gives us 3.29 grams of silver. It's also possible to do all four steps at once in a dimensional analysis approach. So looking at our units, if we multiply 1,225 seconds by 2.40 coulombs per second, seconds cancels out and gives us coulombs. Multiplying that by one mole of electrons per 96,485 coulombs, cancels out coulombs. Next, we have our conversion factor of one mole of silver per one mole of electrons, so moles of electrons cancels out. And multiplying that by the molar mass of 107.87 grams per one mole, moles of silver will cancel out and give us a final answer of 3.29 grams. So we get the same answer, no matter which approach we take.