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Offline zeoblade

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« on: September 04, 2009, 11:15:27 PM »
1) In electrochemistry, cells can be constructed with electrodes of different metals, their solutions, salt bridges all connected in a circuit to measure their voltage. Convention is the Ecathode or Ereduction is stated so that it is always positive. If I measure the Ecell of a metal electrode against a Hg/Hg+ calomel electrode, it is written as positive. Depending on if Hg is the cathode or anode, will the corresponding electrode always be positive? For example:

Ecell = Ecathode - Eanode, where Ecell is measured and Hg/Hg+ = 0.244V. In Ag and Hg system, Ag is being reduced at the cathode and Hg is oxidised at the anode. I measured Ecell = 0.480V so:
Ecathode = Ecell + Eanode by rearrangement
Ecathode = 0.480V + 0.244V = 0.724V, correct?

In Hg and Pb system, Hg is being reduced this time at cathode and Pb oxidised at anode. Measured Ecell = 0.435V so:
Eanode = Ecathode - Ecell by rearrangement
Eanode = 0.244V - 0.435V = -0.191V but this is a negative number. Shouldn't these values always be positive or is it because it is the anode, the value is negative?

2) In Cu and Hg system, Cu is reduced at cathode and Hg oxidised at anode. Does this involve 2e-? Similarly, Hg reduced/Zn oxidised also involves 2e-? The reason for asking is when calculating Eel = Eel0 + 2.303RT/nF x log[Mn+], I'm trying to figure out what value of n to use where n = e- transferrred. For metals Cu, Zn, Ag, and Pb each against Hg/Hg2Cl2 calomel electrode; I can't figure out how many e- are transferred. For each cell where Hg is reduced at cathode, my Eel is negative when I thought it should be positive.

3) I can't find an equation for Gibbs free energy to relate with any of the equations regarding spontaneity. But what I am thinking is, there's a certain limit for the Ecell of a metal that allows it to be spontaneous or not. I can't quite articulate the definition at the moment. There's also a temperature factor and I guess in this case is constant. Temperature is related to enthalpy. In the back of my mind, two equations are coming to me. One is like change in G= -RTlnK or change in G= change in H - change in S x T. But I don't feel any of these are correct because we don't have a K value or an S value. So I'm really lost.

Offline Arctic-Nation

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Re: Electrochemistry
« Reply #1 on: September 05, 2009, 05:33:35 AM »
I'm just starting to learn electrochemistry myself (again), and I'm not going to be able to help you with all your problems, but I'll give you what I know.
In the first place, always try to give full half-reactions. Potentials are dependent on many factors, including counterions. Also, always make clear which ions or species are being investigated. When saying something like 'the Cu/Hg system' you convey absolutely no information about which reactions are to be considered, as Cu to Cu+ to Cu2+ and Cu to Cu2+ (and the reverse ones) are all possible. For Zn it's easier, as it always goes to Zn2+.
Second, in a galvanic (spontaneous) cell, the Eanode, by definition, always has the smallest/more negative potential.
As for your third problem, I don't have a mathematical solution, but for your cell reactions to proceed spontaneously and at a measurable rate, you need a potential difference of at least 0.2 V between the reaction partners, barring any kinetic effects.

Offline zeoblade

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Re: Electrochemistry
« Reply #2 on: September 05, 2009, 06:00:47 AM »
Thank you all the same, Artic-Nation.

The problem is I don't know how many e- are involved in the cell thus can't figure out if its Cu(I) or (II) so I said Cu/Hg system. I guess I'm going to google search it.

With my working out, if you would re-check my calculations, should that be negative values or always positive values?

Offline zeoblade

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Re: Electrochemistry
« Reply #3 on: September 07, 2009, 09:57:40 PM »
Say a situation where Li and Fe are used to create a cell.

The standard reduction potential for Li+(aq) + e- --> Li(s) E0 = -3.040V.

The standard reduction potential for Fe2+(aq) + 2e- --> Fe(s) E0 = -0.44V.

In this cell, Fe is reduced and Li oxidised. The nett reaction is:

Fe2+(aq) + Li(s) --> Fe(s) + Li+(aq).

The number of e- has to be the same so multiply the  Li half reaction by 2 to get:

Li+(aq) + e- --> Li(s) and thus the nett reaction becomes:

Fe2+(aq) + Li(s) --> Fe(s) + Li+(aq).. My question is, because Li half reaction is multiplied by 2, does Li E0 = -3.040V get multiplied by 2 as well?

Offline zeoblade

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Re: Electrochemistry
« Reply #4 on: September 07, 2009, 10:24:37 PM »
As you know, a pH meter performing its function can also be considered an electrochemical cell. A combined glass/calomel indicator electrode for pH measurement is responsive to H+ ions, glass electrodes are sensitive to alkali metal tions in basic solution. For a fixed concentration of Na+ ion, (e.g. 1.0M), in solutions of varying pH, the measured pH is sometimes found to be less than the true pH as the latter is increased. This can be said to be an alkaline error. In addition a typical glass electrode can be said to have an acid error, opposite in sign to the alkaline error in solutions of pH less than about 0.5. The readings tend to be too high. So a glass electrode is being compared to a reference electrode.

In this situation, I have six buffer solutions of increasing pH; 3.1, 4.0, 5.5, 6.5, 7.5, and 9.2. I get corresponding voltages; 0.224V, 0.160V, 0.075V, 0.017V, -0.043V, and -0.136V respectively. I plot pH on x-axis and EMF on y-axis. I obtain a slope of -5.861x10-2 and an intercept of 3.992x10-1. According to the equation:

Ecell = (E0Glass electrode - E0Reference electrode) - 2.303.R.T.pH/F a slight modification to the Nerst equation. Substituting in the values of the constants the equations becomes:

Ecell = (E0Glass electrode - E0Reference electrode) - 1.984x10-4.T.pH, where T is 295K so,

Ecell = (E0Glass electrode - E0Reference electrode) - 5.861x10-2.pH which is the slope that I initially mentioned in this post.

This makes a linear relationship where EMF = y-axis and pH is x-axis. If I were to designated the intercept, please confirm that (E0Glass electrode - E0Reference electrode) would be this intercept.

I would like to find the error in the slope, thus I'm thinking this would be the error in Temperature, i.e., thermometre because -1.984x10-4.T = -5.861x10-2, where T = 295K. Can someone please also confirm this rationale of mine as well?

Offline Borek

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Re: Electrochemistry
« Reply #5 on: September 08, 2009, 02:51:39 AM »
Fe2+(aq) + Li(s) --> Fe(s) + Li+(aq)..

It is not different from the first reaction, you probably forgot coefficients.

My question is, because Li half reaction is multiplied by 2, does Li E0 = -3.040V get multiplied by 2 as well?

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