# Delta and ksp relationship problems

### Calculating the equilibrium constant from the standard cell potential (video) | Khan Academy

What is the value of delta G for the precipitation of AL(OH)3 at 25 C? I then solved for x^2 and plugged it into the equation. . No problem. Click here to check your answer to Practice Problem 5. H - delta (TS). If the reaction is run at constant temperature, this equation can be written as follows. disposal problems. It reinforces the students' understanding of the interrelationship of solubility,. Ksp, ΔG0,. ΔS0 and ΔH0. energy, enthalpy and entropy and the relationships between them. All are incorporated in one. titrimetric experiment.

For our reduction half reaction, we left it how it was written, all right. So we're just gonna write our standard reduction potential as positive. Next, we need to look at our balanced equation, all right. We need to make the number of electrons equal for our half reactions. So for this first half reaction, I'm just gonna draw a line through, I'm just gonna draw a line through this half reaction so we don't get ourselves confused.

For our first half reaction here, we have two electrons.

### Relating delta G to KSP?

Then over here, right, for our oxidation half reaction, we have three electrons. We need to have the same number of electrons. So we need to have six electrons for both half reactions, because remember the electrons that are lost are the same electrons that are gained. So we need to multiply our first half reaction by three. All right, if you multiply our first half reaction by three, we'll end up with six electrons. And our second half reaction, we would need to multiply the oxidation half reaction by two, in order to end up with six electrons.

So let's rewrite our half reactions. So, first we'll do the reduction half reaction, so we have, let me change colors again here. And let's do this color. So we have three times, since we have three I two now, We have three I two, and three times two gives us six electrons. So three I two plus six electrons, and then three times two, all right, gives us, three times two gives us six I minus.

All right, so we multiplied our half reaction by three, but remember, we don't multiply the voltage by three, 'cause voltage is an intense of property. So the standard reduction potential is still positive. So we have positive. Next, we need to multiply our oxidation half reaction by two.

So we have two Al, so this is our oxidation half reaction, so two Al, so two aluminum, and then we have two Al three plus, so two Al three plus, and then two times three gives us six electrons.

So now we have our six electrons and once again, we do not multiply our standard oxidation potential by two, so we leave that, so the standard oxidation potential is still positive 1. Next we add our two half reactions together. And if we did everything right, we should get back our overall equation.

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• Calculating the equilibrium constant from the standard cell potential
• Standard change in free energy and the equilibrium constant

So our overall equation here. We have six electrons on the reactant side, six electrons on the product side, so the electrons cancel out. And so we have for our reactants three I two, so we have three I two, plus two Al, and for our products right here, we have six I minus, so six I minus, plus two Al three plus, plus two Al three plus.

So this should be our overall reaction, all right, this should be the overall reaction that we were given in our problem. Let's double check that real fast. So three I two plus two Al, all right, so right up here, so three I two plus two Al, should give us six I minus plus two Al three plus. So six I minus plus two Al three plus. So we got back our original reaction.

Remember our goal was, our goal was to find the standard cell potential E zero, because from E zero we can calculate the equilibrium constant K. So we know how to do that, again, from an earlier video. To find the standard cell potential, all right, so to find the standard cell potential, all we have to do is add our standard reduction potential and our standard oxidation potential.

## 16.3: Precipitation and the Solubility Product

So if we add our standard reduction potential and our standard oxidation potential, we'll get the standard cell potential. So that would be positive. So the standard potential for the cell, so E zero cell is equal to. All right, now that we've found the standard cell potential, we can calculate the equilibrium constant.

### Gibbs Free Energy

So we can use one of the equations we talked about in the last video that relates the standard cell potential to the equilibrium constant. So I'm gonna choose, I'm gonna choose one of those forms, so E zero is equal to 0.

So again, this is from the previous video. The free energy of a reaction at any moment in time is therefore said to be a measure of the energy available to do work.

When a reaction leaves the standard state because of a change in the ratio of the concentrations of the products to the reactants, we have to describe the system in terms of non-standard-state free energies of reaction.

The difference between Go and G for a reaction is important. There is only one value of Go for a reaction at a given temperature, but there are an infinite number of possible values of G.

The figure below shows the relationship between G for the following reaction and the logarithm to the base e of the reaction quotient for the reaction between N2 and H2 to form NH3. They therefore describe systems in which there is far more reactant than product.

The sign of G for these systems is negative and the magnitude of G is large.

The system is therefore relatively far from equilibrium and the reaction must shift to the right to reach equilibrium. Data on the far right side of this figure describe systems in which there is more product than reactant. The sign of G is now positive and the magnitude of G is moderately large. The sign of G tells us that the reaction would have to shift to the left to reach equilibrium.

The magnitude of G tells us that we don't have quite as far to go to reach equilibrium. The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram are particularly important. The straight line crosses the vertical axis when the reaction quotient for the system is equal to 1.

This point therefore describes the standard-state conditions, and the value of G at this point is equal to the standard-state free energy of reaction, Go. Because there is no driving force behind the reaction, the system must be at equilibrium.