## Calculating ln Q

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Bryce Ramirez 1J
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### Calculating ln Q

Can someone explain what Q is and how to use it in the Nernst equation?

chemboi
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### Re: Calculating ln Q

Q is [product]/[reactant], use in the appropriate place in the expression for the Nernst equation (will be given on eq. sheet)

Anthony Hatashita 4H
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### Re: Calculating ln Q

Q is the concentration of product divided by the concentration of reactant. [product]/[reactant]. Lots of questions will give you the concentrations and you'll be able to solve from there.

Bryce Ramirez 1J
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### Re: Calculating ln Q

Anthony Hatashita 4H wrote:Q is the concentration of product divided by the concentration of reactant. [product]/[reactant]. Lots of questions will give you the concentrations and you'll be able to solve from there.

When you say product/ reactant, do you mean that you take the part on the right of the cell diagram and divide that by the part on the left of the cell diagram?

805312064
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### Re: Calculating ln Q

Yes because the right side is the cathode (gaining electrons) while the left is the anode (losing electrons).

Jessica Li 4F
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### Re: Calculating ln Q

I find it helpful to write out the cathode and anode reactions, even though they are roughly the same.

For instance, the half reactions would be:
Anode: Metal -> Metal2+(aq, anode conc.) + 2e-
Cathode: Metal2+(aq, cathode conc.) + 2e- -> Metal

If you add them up, you can easily get the overall equation and calculate Q.

805373590
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### Re: Calculating ln Q

Q is the concentration of product over reactant in other words for galvanic cells we can write q as the concentration of anode over that of the cathode.

Nathan Nakaguchi 1G
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### Re: Calculating ln Q

Q is the reaction quotient, [Products]/[Reactants] for concentration cells the lower concentration is the products. Or the cathode concentration is the reactants and the anode concentration is the products.

Verity Lai 2K
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### Re: Calculating ln Q

Q is just the reaction quotient which is from the equilibrium portion of 14b. You use it to connect the molarity of the two solutions with the cell potential.

Nuoya Jiang
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### Re: Calculating ln Q

Q is the reaction quotient. When we are using the Nernst Equation in a concentration cell, we should try to make lnQ negative.

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### Re: Calculating ln Q

Nathan Nakaguchi 1G wrote:Q is the reaction quotient, [Products]/[Reactants] for concentration cells the lower concentration is the products. Or the cathode concentration is the reactants and the anode concentration is the products.

Why is the cathode concentration the reactant if the the cell diagram shows it in the order of the anode to the cathode?

Sartaj Bal 1J
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### Re: Calculating ln Q

The cathode concentration is considered the reactant because we were told to make the lower concentration out of the two the product. Usually, the anode has a lower concentration in a concentration cell, making it the product.

Posts: 125
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### Re: Calculating ln Q

Bryce Ramirez 1J wrote:Can someone explain what Q is and how to use it in the Nernst equation?

Q is the equal to the concentration of products over reactants. In the concentration cells, if you find the two half reactions, you will find that Q is equal to the concentration of the anode, over the concentration of the anode.

faithkim1L
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Joined: Fri Aug 09, 2019 12:17 am

### Re: Calculating ln Q

Q is the concentration of products divided by the concentration of reactants. Put the coefficient as the exponent for the concentrations of products and reactants.

HuyHa_2H
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Joined: Wed Sep 11, 2019 12:15 am

### Re: Calculating ln Q

Q is the ratio of concentrations and/or partial pressures of products over reactants where the individual coefficients are to be used as exponents for each individual reactant/product. You can use Q to find other values of variables that are also on the Nernst equation if given enough additional information.

Rebekah Alfred 1J
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### Re: Calculating ln Q

When calculating Q for the Nernst equation, you can use both partial pressure and concentration. Before, all the values plugged into Q had to be all concentration values or partial pressure values.

Jasmine Vallarta 2L
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### Re: Calculating ln Q

If rR --> pP, then Q is the ([P]^p)/([R]/^r) at any given time of the rxn; Q is different from Kc, the equilibrium constant; Q can be found with concentrations or partial pressures

Sean Cheah 1E
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### Re: Calculating ln Q

Q is easy enough to find for typical redox reactions. It gets a little weirder for concentration cells, but writing out the half-reactions should immediately make it clear which concentration to use for the products and which to use for the reactants. What I don't understand is how one is supposed to come up with a convincing expression for Q in cells involving pH meters. For instance, the official answers to 6.63 and 6.65 (both problems involving pH meters) use different expressions for Q ([H+] in the first case and [H+]/[OH-] for the latter). Can someone explain how they came up with these Q expressions and why they are different?

Part of the problem is that I simply don't really understand the mechanism behind the combination electrode that most pH meters use today. I've been scouring the web for days now but still have yet to find a single source offering a logical explanation for what is going on inside these devices in the traditional redox sense. I understand that a difference in the degree of the binding of protons (and the displacement of metal ions in the hydrated layer) on either side of the ion-selective glass membrane results in a potential difference. The pH meter is calibrated in such a way so that the constant (and irrelevant) potentials created by the Ag/AgCl electrodes are factored out, leaving only the relevant potential caused by the aforementioned difference in protonation between the two sides of the glass (one side in contact with the electrode's inner buffer solution of known pH and the other with the outer test solution of unknown pH). How exactly is one supposed to write a traditional redox equation for this process though? The best I could do is write out the half-reactions for the Ag+/Ag half-cells occurring at each metal electrode, but this entirely ignores the all-important "third and fourth" electrodes formed by the two sides of the ion-selective glass. This is problematic because the potential that we care about, the one that is pH-dependent, occurs between the two sides of the glass membrane. Without a balanced overall redox equation, how do I know which species to treat as products, which to treat as reactants, and what exponent to raise each to in my reaction quotient Q?

Shail Avasthi 2C
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### Re: Calculating ln Q

The ion in aqueous solution that is growing in concentration is the product (i.e. at the anode). The ion in aqueous that is decreasing in solution (i.e. at the cathode) is the reactant. The pure metal at the anode and cathode is not involved in Q because they are solids. Solids and solvents don't factor into Q.

Ryan_K_1K
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### Re: Calculating ln Q

Q is the concentration of the products of your balanced reaction divided by the concentration of the reactants. It is important to note that the concentrations are raised to the power of the coefficient value in front of a species.

J Medina 2I
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Joined: Wed Sep 25, 2019 12:17 am

### Re: Calculating ln Q

Q is the the concentration of the products divided by the concentration of the reactants. It is used in the Nernst equation as a replacement for K when a reaction is not at equilibrium.

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