# Thread: Help! Cyclic Voltammetry and electrodeposition

1. Hello,

I wish to deposit tin onto a carbon rotating disc electrode to use it to produce formate from a co2 solution later on.
I'm thinking of using a 1 mM tin (II) chloride in 30% acetonitrile, 70% toluene solution. I want to control the tin thickness, googling found me a paper saying that they calculated the amount of voltage need applied and the time required through cyclic voltammetry but didnt say how, so I googled that too and......nothing, or at least some equations i don't understand enough to apply. Can somebody point me to the right direction, or better describe how can one adjust the thickness of the tin deposits with CV? It will be very much appreciated thank you.  2.

3. Thanks for the link, though I only can access a cache copy of it.

I shall quote the passage which I had referred to below;

'After the experiments had been conducted on the vitreous carbon electrode it was then necessary to investigate the effects of depositing Tin onto the electrode surface. In order to do this a further cyclic voltammetry experiment were conducted on a solution of 0.067M SnO2 to see what voltage needed to be applied in order to reduce the Sn4+ in the solution into Sn (solid). The voltage that was decided to be applied was -0.69V (SCE). After this, calculations were carried out based on the cyclic voltammtery experiments in order to find the length of time this voltage needed to be applied for in order to produce a 10-6m (1µm) deposit on the surface of the electrode. This length was decided as it was well within the Nernst diffusion layer for the reaction. The length of time that was calculated was 467 seconds. In order to do this an electrochemical cell was set up using a catholyte of a 0.067M SnO2 and an anolyte of 1.00M H2SO4, a platinum flag electrode was the counter electrode used. After the Tin had been deposited a rotation speed of 800rpm was used to compare the current density for CO2 reduction to that of the vitreous carbon disk electrode. In order to do this the same method as described above for the vitreous carbon experimentation was used.'

They did not say that they controlled the deposition THROUGH CV, but rather, based on 'calculations were carried out based on the cyclic voltammtery experiments' done beforehand. I hope it's clearer now, sorry!  4. Thanks for the insight, so I gathered from your explanation,

First calculate the current that would be supplied at that voltage, I =V/R

Also, I = (ne)/t

Then number of electrons, n = (It)/e

Then total number of Sn4+ molecules reduced = (It)/4e

Then total moles of Sn4+ reduced = (It)/4e(Avagadro No)

Then total grams of Sn4+ reduced = (It)/4e(Avagadro No)(MW)

Then total volume of Sn4+ reduced = (It)/4e(Avagadro No)(MW)(density of tin)

Then thickness of tin deposited = total volume of Sn4+ reduced/ area where area and the thickness is known, then solve for t,

Is it correct?  5. ....and another thing, the value -0.69V you mentioned as the reduction potential wasn't stated as such in the passage, but as a voltage they decided to apply. Also, the calculations above did not take account the molarity of the SnO2 solution which I imagine it to have an effect. How do you suppose that they came out with the number 0.067M of SnO2 ? It looks to me like it is a calculated value.  6. Hmm, I'll read up on Nernst first, thanks!  7. Okay...sorry to intrude further, but if E is made to be spontaneous for the cell, wouldn't the reaction happen everywhere in the cell?
If I intend for it to happen only on the surface of my working electrode, do I adjust the concentrations so that it is still not spontaneous anywhere except at the surface of my electrode where I apply a potential to make up the difference?  8. Okay, thanks again, have to ponder your words a bit.  9. I have a question about the meaning of the nernst potential.

say for a reaction A + B = C + D with a caculated nernst potential of -0.5

Does it mean that if I apply a potential of -0.4, the reaction will move forward, i.e. produce C and D?  10. so... it means i have to apply a potential > +0.5V to make the Gibbs energy negative and hence the reaction spontaneous?  11. well okay, you've been a great help. Thanks alot.
I don't actually study chemistry in uni, but recently I have to undertake a project in uni which involves all these stuff, and they are pretty new to me. Have to learn it by myself and hence the stupid questions.
At any rate, thanks again.  Bookmarks
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