Electrolysis, charge of ions.

[AdamCFC]

i do but i dont, i have the initial currents for all three. i don't have the end current for the +3, and theyre all over the place anyway. What do you think of my graph, and the gradient analysis?

 

[Bystander]

Without the current data, it is meaningless.

How much of the definitions given for "I = nAve" is from material given to you with the assignment, and how much is paraphrase? There are some misunderstandings that are killing you, and we've got to get them identified.

 

[AdamCFC]

I only have initial current readings, that's all really. They were all 0.65A. But i didnt use currents to work out moles deposited. I am measuring moles deposited, and i found that out by dividing the mass change by atomic mass.

 

[Bystander]

Let's look at something else a bit --- try a different tack.

Compare the sequence of mass changes for each series of three runs of Ag and Cu to that for Fe. Anything strike you as interesting?

 

[AdamCFC]

In my view, the trend should be linear. Look at +2, if we took the average to be the top outlier, there would be a linear fit/trend. It wouldn't be prefect but my results arnt completely precise. This would also make the trend directly proportional, this is what i would have expected, but i can't explain why.

This begs the question, that maybe 0.001102 ( the outlier ) is infact the real answer, and 0.000670 and 0.000472 are outliers, its plausible and would make sense. I will include this in detail in my evaluation OR conclusion

Adz

 

[Bystander]

(snip) there would be a linear fit/trend. It wouldn't be prefect but my results arnt completely precise. This would also make the trend directly proportional, this is what i would have expected, but i can't explain why.(snip)

Ah-hah! Now, you take advantage of odd little observations made during the experiments. For instance, looking just at the three runs with Ag, what can you tell me about them? Did you use the same pair of electrodes for all three? Did you notice anything about the metal plated onto the electrode? (Coverage, bare spots, that sort of thing) And, just a general question, what can you tell me about the surface area of the rough side of a piece of sandpaper compared to the "smooth" side?

 

[AdamCFC]

I can't remember much, this was 6 weeks ago. I remember that the we did use the same electrodes. But after the first run there's was a strong coverage, the other 2 runs wernt patch as such, but there was a difference. I can only think an error with the molarity, ditance, voltage, or change in the variable resistor caused this massive difference.

The rough side has a far greater surface area, just like the small instestine, the fingerlike bumps (villi) cause the surface area to greatly increase.

 

[Bystander]

... , and, if you paint the rough side of the sandpaper with a gloss enamel?

 

[AdamCFC]

wow that's a great point :) we did use a cloth which did get rid of a lot of deposited metal, and we then rinsed it in acetone.

Would you have expected the trend to be linear?

 

[Bystander]

"Linear?" If I can control n, A, v, to be constants among the various e values, yes; I know I can't, so, the most I can expect is an approximation of linearity.

Let's review:
How are you controlling n, constant, and were you successful?

How A, and success?

How v, and success?

 

[AdamCFC]

Number of Ions - Kept constant by using same amount of electroyte at same molarity (concentration). - Yes i believe it was successful

Surface area of electrodes - Kept constant by using same electrodes, and made sure the same depth was maintained in the solutions - 50% successful, the surface area decreased after each test. The plating process was "smoothing out" minute imperfections in the electrode, decreasing the surface area, thus decreasing the amount of electrode in contact with the solution

Velocity of Ions -Kept constant by using the same power-pack at 12V, and eliminating errors by using the same power-pack throughout my experiment, doing all 3 tests during a 3 day period, at room temperature. ? Keeping the Amps the same by using a variable resostor ? - Yes i think this was successul, it wasnt 100% precise, but i sucessfully did what i could with the resources i had.

Why should the trend be linear?

 

[Bystander]

"n:" decreased through each run as material is plated on the cathode; changed from Ag to Cu to Fe as number of anions increased.

"A:" increases or decreases as material is added to the cathode (forming dendritic growths to increase, filling pores to decrease), but cannot remain constant. Compare the Ag and Cu results from 1 to 3.

"v:" no current measurements, no way of knowing much of anything about anion current; switching anions from nitrate to chloride; changing cation masses; changing cation radii (sizes); and none of this is even going to be on your radar until AFTER you've been through your first chem course.

"Why linear?" If n, A, v are constant, the expression I = k times e is linear in e.

 

[AdamCFC]

Are you saying my analysis of n A and v are wrong? or are you adding to them?

"n:" decreased through each run as material is plated on the cathode; changed from Ag to Cu to Fe as number of anions increased

Isnt this charge, not number, as i am changing ion charge?

Yes, by comparing my results from 1-3 on Gold and Copper, i see the mass changes decrease slightly over the 3 runs. Wouldnt the "growths" and "fill ins" roughly equal up? or do they have tendencies to do just one?

Velocity, didnt i control these well, both through temperature and input voltage?

Finally, I = k x e - what is k?

Ive heard k is "so many times" something? e.g k X 9 would be the 9 times table?

Thanks

 

[Bystander]

Are you saying my analysis of n A and v are wrong? or are you adding to them?



Isnt this charge, not number, as i am changing ion charge?

"n" is the concentration of ions per unit volume; "e" is the charge of a particular ion. 0.1m AgNO₃ ionizes when dissolved in water to give you 0.1m NO₃^-1 and 0.1m Ag⁺¹ which is not the same value of "n" for 0.1m FeCl₃, 0.1m Fe⁺³ and 0.3m Cl^-1. The current carrying product of n and e for the two salts is different, 0.2 for Ag and 0.6 for Fe.

Yes, by comparing my results from 1-3 on Gold and Copper, i see the mass changes decrease slightly over the 3 runs. Wouldnt the "growths" and "fill ins" roughly equal up? or do they have tendencies to do just one?

This is what is termed "throw," and is different for every metal and for every plating solution/recipe. Some metals cover the cathode more uniformly than others. You can get smooth plating films, rough blobs and gobs, metals that grow staghorns and trees. It's part of the "art" that is necessary to make a living running a plating shop.

Velocity, didnt i control these well, both through temperature and input voltage?

Think of what impedes the motion of ions through solution, and what forces are at work to move the ions through solution. Voltage between the electrodes? Nope, and you ain't responsible for knowing any of this --- isn't even seen by ions in aqueous solution. What makes 'em move? The concentration gradient; when an ion is reduced at the cathode, the local concentration drops, and ions from more concentrated regions diffuse into the region of lowered concentration. The diffusion is a function of temperature, "good on you" for knowing to run all your experiments at as nearly the same temp as possible.

Don't sweat what you haven't learned yet.

Finally, I = k x e - what is k?

"k" is a constant, or presumably a constant, the product of n, A, and v in your expression for current. Since you increased both n and e, you'll want to look at just k = Av as being held constant.

Ive heard k is "so many times" something? e.g k X 9 would be the 9 times table?

Thanks

 

[AdamCFC]

"n" is the concentration of ions per unit volume; "e" is the charge of a particular ion. 0.1m AgNO3 ionizes when dissolved in water to give you 0.1m NO3-1 and 0.1m Ag+1 which is not the same value of "n" for 0.1m FeCl3, 0.1m Fe+3 and 0.3m Cl-1. The current carrying product of n and e for the two salts is different, 0.2 for Ag and 0.6 for Fe.

:/ i don't understand much of this.

So, I = k x e ... = ... I = ( n x A x v ) x e

so because i just increased e in equal increments, that's why the line/trend should be linear. but my slight errors in keeping n constant have caused it to become a curve?

Can i include this, and that equation in my evaluation/conclusion?

and what did u think about the start of my analysis on the last page?

:)

 

[AdamCFC]

Just an extra thing to my last post. I would like to include a hypothesis, does this sound suitable and where would i put it?

If moles deposited is related to ion charge, then increasing the charge will increase the current and mass change, which in turn will increase the amount of metal deposited.

Do the bits in bold need to be re-arranged?

 

[Bystander]

You have written "I" as the product of four terms. If you hold three constant and change 1, "I" is linear in that one term. If you hold two constant, and change the product of two others, "I" is linear in the product of those two terms. If you hold one constant, change the product of two others, and have no control over the third, you're in trouble.

"Mass" or "moles?" Both start with "m," but they are NOT interchangeable.

You can discuss crumple zones --- you can count charge carriers, the product of charge carrier concentration and charge --- you can distinguish between mass and mole.

 

[AdamCFC]

Was what i said about crumple zones right, id feel terrible if i was giving someone false information.

Okay, i think that's fine now :) My intention was to control n A and v and just change e. That would have given me a linear trend. Due to errors and things i couldn't control precisely (temperature), i only controlled ...

wait, i havnt controlled/kept constant anything?

n - didnt 100% understand what you said about ion concentration per unit volume, but i understand that by using 100ml at each at 0.1mol, that was not enough to control ion concentration, and i did not control this well.

A - The surface area varied with each run

This is what is termed "throw," and is different for every metal and for every plating solution/recipe. Some metals cover the cathode more uniformly than others. You can get smooth plating films, rough blobs and gobs, metals that grow staghorns and trees. It's part of the "art" that is necessary to make a living running a plating shop.

v - Volts - Kept constant
-- Temperature - Did vary day by day, not much, but there was a change

This is not a total diaster ... as far as schoolwork goes, because.
I have a trend that supports my prediction and hypothesis
I have a lot of errors i can talk about, and most importantly, describe how to minimise/eradicate in future experiments. e.g, do all tests at a controlled temperature.

 

[Bystander]

Was what i said about crumple zones right, id feel terrible if i was giving someone false information.

Wasn't wrong. You gave a hint, pointed to a source --- abided by the "don't do their work for them, get them to do it themselves" spirit of the homework forum.

I was pointing out to you that you are NOT a babe in the woods --- that stuff's been covered for you and basic concepts relevant to this electrolysis problem haven't been. Punched the frustration button there for me.

Okay, i think that's fine now :) My intention was to control n A and v and just change e. That would have given me a linear trend. Due to errors and things i couldn't control precisely (temperature), i only controlled ...

wait, i havnt controlled/kept constant anything?

n - didnt 100% understand what you said about ion concentration per unit volume, but i understand that by using 100ml at each at 0.1mol, that was not enough to control ion concentration, and i did not control this well.

A - The surface area varied with each run

v - Volts - Kept constant
-- Temperature - Did vary day by day, not much, but there was a change

This is not a total diaster ... as far as schoolwork goes, because.
I have a trend that supports my prediction and hypothesis
I have a lot of errors i can talk about, and most importantly, describe how to minimise/eradicate in future experiments. e.g, do all tests at a controlled temperature.

So long as you learn, there are NO "total disasters." Good.

You predict a "linear trend." Let's work on that, because there is something in "I=nAve" that is still leading you off into the weeds.

Looking just at silver and iron, +1 and +3 charges, you expect to see the current triple; you are going to "measure" current by weighing the cathode of each cell before and after a timed electrolysis and calculating moles plated onto the cathode. What is your prediction for the number of moles of iron? Less than, greater than, or equal to the number of moles of silver?

 

[AdamCFC]

er, I am not sure. I don't think i expect iron+3 to have three times more moles deposited than silver +1. I do expect a proportional relationship though. I expect more moles deposited on the cathode after using the iron electrolyte than the silver electrolyte. This is because iron ions have a higher charge.

Am i right in thinking, because silver has almost double the atomic mass of iron, that 10 iron ions would need to be plated to equal 5 silver ions?
If this is the case, because iron deposited a higher mass change than silver, many many more ions were deposited on the cathode than the amount of silver ions deposited.

Going on holiday wednesday morning, so won't be able to reply after that until the 1st sept.

Adam

 

[Bystander]

er, I am not sure. I don't think i expect iron+3 to have three times more moles deposited than silver +1. I do expect a proportional relationship though. I expect more moles deposited on the cathode after using the iron electrolyte than the silver electrolyte. This is because iron ions have a higher charge.

Wrong. You have increased the current by increasing the charge on the ion. You have not increased the number of ions being deposited. Silver at +1 carries a current I; iron at +3 carries a current of 3I. A silver ion will be reduced by ONE electron to atomic (metallic) silver at the cathode; a ferric ion requires THREE electrons to be reduced to metallic iron. Recheck the definition of electric current.

Am i right in thinking, because silver has almost double the atomic mass of iron, that 10 iron ions would need to be plated to equal 5 silver ions?

You are determining mass only as a means to find the numbers of moles deposited; other than for that purpose, mass is meaningless in the context of this experiment.

If this is the case, because iron deposited a higher mass change than silver, many many more ions were deposited on the cathode than the amount of silver ions deposited.

Yes. But, please think in terms of moles rather than mass. You want to count charge, you want to count moles, you want to compare those numbers rather than masses.

Going on holiday wednesday morning, so won't be able to reply after that until the 1st sept.

Adam

 

[AdamCFC]

Ok, so to your previous question, I predict there will be more moles of iron deposited than moles of silver.

there is something in "I=nAve" that is still leading you off into the weeds.

Whats that?

 

[Bystander]

Silver ion at +1, current is I; ferric ion at +3, current is 3I. WHAT is the definition of current? Same time for both cells, total charge transferred is what for silver? And what for iron? Reducing how many moles of silver? And how many moles of iron?

You've got two guesses left.

 

[AdamCFC]

Current represents the flow of electrons through a conductive material.

Silver needs 1 electron to become an atom again, Iron needs 3.

 

[Bystander]

Current is defined as CHARGE per unit time, be that charge carried by electrons, or ionized atoms or molecules.

Now, more moles of iron than silver? Same? Fewer?

 

[AdamCFC]

More moles of iron than silver.
More mass deposited using iron than silver.
Higher current using iron than silver.

 

[Bystander]

Wrong.
Wrong.
Right.

Charge is defined as the sum of the products of the numbers of charge carrying species with the numerical values of the charges on the species.

Try again.

 

[AdamCFC]

sorry this is over my head, its just guesses now. looking at my results, iron deposited more moles than silver and copper.

 

[Bystander]

We're working on your prediction which you are then going to compare to the results.

If cations are the only charge carrying species in solution, silver at +1, or iron at +3, at the same concentration, you predict three times the current. If you weigh the cathodes, divide by atomic weight to get number of moles, and compare the number of moles, is the number of moles of iron less than, equal to, or greater than the number of moles of silver?

 

[AdamCFC]

Silver - 0.033g / 108 = 0.000306moles

Iron - 0.110g / 56 = 0.001964moles

The number of moles of iron(1964) is greater than the number of moles for silver(306).

 

[Bystander]

I'm not even going to ask where those numbers came from.

One silver ion is reduced in time t; one ferric ion is reduced in the same time t. For silver, the current is one unit charge over t; for iron the current is 3 unit charges in the same time t, or three times greater current. The number of moles of silver is one over Avogadro's number; the number of moles of iron is ONE over Avogadro's number.

 

[AdamCFC]

those were the exact numbers out of my coursework table, look at the moles deposited table.