Cell Potential: ∆G and K - Video Tutorials & Practice Problems

Relationship between ∆Eº, ∆G and K

Now recall that when we talk about a spontaneous reaction, it will possess a standard cell potential that is greater than zero, a change in her standard gives free energy that's less than zero and an equilibrium constant decay, that's greater than one. Now, because we can talk about the conditions to make up reactions spontaneous with these variables. That means we can connect them to one another through equations. Here, we're gonna say the relationship between these three variables and spontaneity can be observed under the following diagram. So here we have our triangle diagram which will connect our three variables together. So let's start on the left side and we're looking at the equation that connects together our equilibrium constant K and our standard cell potential. So here we'd say that our standard cell potential equals RT over N times F times Ln of K. Here we're gonna say that ee sub not, well, eno sub cell equals our standard cell potential. Uh We would say uh delta G change in our standard gives free energy is gives free energy, artur constant is K and here is just the moles of electrons transferred. We'd say F here is far constant in 96,004, 85 coulombs per mole of electrons. So that's 485 and R here is our gas constant, which is 8.314 joules over most times K. Now let's move clockwise around this triangle. So now we're going over this way and we're seeing the connection between our standards of C potential and gibbs free energy. So here we'd say that change, instead, it gives free energy equals negative N times F times our standard cell potential. So it's equal to the negative N which is moles of electrons transferred times Friday is constant times our standard cell potential. Let's keep going clockwise. Finally, we look at the, the last equation that connects our equilibrium constant K and gives free energy. Here, we would say that our equilibrium constant K equals E to the negative delta G not divided by RT. So we realize that these three variables can help us determine if a reaction is spontaneous because we can relate them to each other. In terms of this um qualitative analysis, we can look at their equations and how they connect um in terms of quantity to one another. So keep in mind the different formulas that connect together these three types of variables.

Cell Potential: G and K Example

Here it says calculate the standard cell potential for the following reaction if 10 moles of electrons are transferred. So here we have P four solid reacting with five moles of oxygen gas to produce one mole of tetro phosphorus Decca oxide. So here we're going to say they're giving us gibbs free energies of formations for each one of these compounds. Remember if an element is found in its natural state, its gibbs free energy of formation is equal to zero. We're gonna say here that change in standard gives free energy equals negative N times Friday's constant stock times our standard cell potential. We're told that 10 moles of electrons are transferred. So N is 10 moles of electrons. F is far is constant which is 96,485 cool lumps per moles of electrons. We're looking for our standard cell potential. So this is our unknown variable. Before we can find it, we also need to know that change in our standard gives free energy. How exactly do we find that? Well, since they're giving us, gives free energy of formation values, we can say that the change in the standard gives free energy of my reaction equals products minus reactants. So we'd say that is one mole of P 4010 with each one being negative 2984 kilojoules per month minus my reactant to both zero. So we don't even need to include them. So moles cancel out. So I have negative 2984 kilojoules. What we need to remember here is that when it comes to our standard cell potential, it's in units of volts, but a vault is equal to jewels per cooler. So that means that I need to convert these killer Jes into jolts. 1 kg jaw is equal to 10 to the three jaws. So that comes out to be negative 2984000 tools. What we do now, moles of electrons cancel out, divide out negative 10 times 96,045 coulombs from both sides. So 96 485 coulombs. So this all cancels on the right side. And at the end, what are we gonna have? We're gonna have joules per Coolum, which is equal to volts. So our standard cell potential here will come out to be 3.09 volts. So this would be our final answer.

Given the following standard reduction potentials, determine Ksp for Hg₂Cl₂₍s) at 25 °C.

Hg₂²⁺ (aq) + 2 e^– → 2 Hg (l)                           E°red = + 0.789 V

Hg₂Cl₂ (s) + 2 e^– → 2 Hg (l) + 2 Cl ^– (aq)                E°red = + 0.271 V

• A. 4.93 x 10¹¹
• B. 2.18 x 10¹⁶
• C. 3.27 x 10¹⁷
• D. 6.74 x 10⁴

What is the value of the cell potential for the 4 electron transfer reaction below if the equilibrium mixture contains 0.255 M of CH₄, 1.10 M CO₂, 0.388 M CO and 0.250 M H₂ at 25ºC?

                                     CH₄ (g) + CO₂ (g)  ⇌. 2 CO (g) + 2 H₂ (g)

• A. +1.2720 V
• B. -0.2810 V
• C. +0.3021 V
• D. -0.0218 V

Given the reaction: 2 Cl2 (g) + 2 H2O (g) ⇌ 4 HCl (g) + O2 (g) Kp = 7.5x10-2, calculate the Gibbs Free Energy change for the reaction below at 30ºC.

8 HCl (g) + 2 O2 (g) ⇌ 4 Cl2 (g) + 4 H2O (g)

• A. -1.3 x 104 J/mol
• B. -2.9 x 103 J/mol
• C. +4.3 x 105 J/mol
• D. +7.7 x 107 J/mol