Consider the reaction shown here occurring at 25°C. Cr(s) + Cd²⁺(aq) → Cr²⁺(aq) + Cd(s) Determine E°_cell, K, and ∆G°_rxn for the reaction and complete the table.
[Cd²⁺] [Cr²⁺] Q E_cell 𝚫G_rxn
1.00 1.00
1.00 1.00 × 10^-5
1.00 × 10^-5 1.00
4.18 × 10^-4 1.00

Question

Consider the reaction shown here occurring at 25°C. Cr(s) + Cd²⁺(aq) → Cr²⁺(aq) + Cd(s) Determine E°_cell, K, and ∆G°_rxn for the reaction and complete the table.

[Cd²⁺] [Cr²⁺] Q E_cell 𝚫G_rxn

1.00 1.00

1.00 1.00 × 10^-5

1.00 × 10^-5 1.00

4.18 × 10^-4 1.00

Table showing concentrations and values for electrochemical cell calculations in electrochemistry.

Accepted Answer

Hello everyone today, we have been given the following question for this following reaction here at 298°. Kelvin, calculate the cell potential B K and the gibbs free energy of the reaction and fill out the table below. So first, you want to split these into two half reactions. So we're gonna have a cathode and an end node at our cathode, we're going to have our mercury two plus Being given two electrons to form our liquid mercury. And at our an ode, we're going to have our two moles of iodide to form our iodine solid plus two electrons. And this is just based on the fact that one is positively charged and one is negatively charged. And so if we were to look this up in a reference text, the Sell potential for reduction is going to be 0.85, And the cell potential for our reduction for this iodide is going to be 0.536V. And so from that, we can calculate our self potential. So our cell potential is going to be our potential for the reduction at the cathode minus the self potential for reduction at the anodes. And so that's going to translate as 0.851 volts minus 0.536 volts to give us 0.3 volts or . jewels per Coolum. And so with that, we can solve our gibbs free energy for the reaction. So our gibbs free energy for the reaction is going to be negative, the molds that we have times our faraday's constant times our cell potential. And that's going to translate to we have negative and we have two electrons. Faraday's constant can be found in the reference text is 96485 coolants per electron. And then our self potential we just calculated was .315 jules pour coolant. And so that's gonna give us negative 60,785 jewels. However, we need to have this in units of killer jewels. So we'll just use the conversion factor that one kg jewels equal to 10 to the third jewels. This will give us a final answer for that to be negative 60.8 kg joules. And then of course, we are going to solve for our K before we get into solving for the table. So to solve for K, what we need to do is we need to have our cell potential and we're gonna have that formula that is equal to and this is a constant zero .059, 2 volts per moles. We're gonna multiply that by log of K, then we're going to rearrange this equation such that log of K, it's equal to our moles Times our self potential divided by our .059, And of course, when we do this, we're going to plug in our two electrons as we did before, Our cell potential was .315V or .315 jewels per Coolum. And then we're going to divide by that .05, 9, 2V. And this is going to lead us with an answer of 10.64. However, we still have that log of K. So to solve for K potentially where to take 10 and raise it to that value that we had, which was 10.64. And this will give us K which is 4.38 times 10 to the 10th. So we have solved for our gifts for energy, our K value and our self potential. And so now we need to solve for our table. So we're gonna break this up into rows. We're gonna have our Rome one here. And the first thing we saw for is for our Q. So our Q is going to be equal to the products over the reactant. And so we know that there that the only reactant, we have our, our, our mercury and our iodine right here and are here respectively. But we don't include liquids and solids. So this is essentially just going to be one over the concentration of our iodine and our mercury. So that's basically going to be our 0.5 0.1. This is gonna be squared for the, for the iodine because iodine has a coefficient of two. And then we're gonna multiply that by the coefficient Or by our mercury concentration, which is also .1. And when we plug into a calculator, we're going to get one 1000. So our q value for our first row is 1000, never gonna calculate for the cell potential. And that's gonna be equal to or just the electric electrochemical potential that's going to be equal to the cell potential minus our 0.592 volts per mole times our log of Q. And so that's going to be our zero 0.3, 15 volts minus our 150. volts for volts per moles, which is two times our log of Q, which we calculated was 1000, giving us a value of 0.2262 volts. So we have 0.2262 volts. They're gonna calculate four hour gives for energy for the reaction. And that's going to be equal to our earlier gibbs free energy plus our gas constant times temperature times our natural log of Q. And so we calculated that earlier to be negative 60. kg joules Plus our gas constant was 8. Times 10 to the negative 3rd kilo joules per mole. And our temperature was 2 98 Kelvin. And then we had a natural log of 1000 as our queue. And this is gonna be a value of negative 0.7 kg jules. We're gonna have negative 43.7 kg Joel's And we're gonna apply the same concept to our 2nd row. So we're going to have row to here for our road to here, we're gonna have our Q value like we said, that was products over reactant. However, we only, we don't have any products. So our reactant is just going to be the concentration of our iodine, Which will be squared once again times our concentration of our Mercury two plus. And so that was going to be one over and our concentration of ID for the second row was . and our concentration For our Mercury was three times to the -3. So we plug that into the calculator. We get a value of 1300 and 33.3, 1333.33. And it will be calculated for our electro chemical potential to sell. It's going to be our previous cell potential minus our 0.592 volts per moles times our log of Q. So that's going to be our 0.3, minus our constant here Times our Lagos que which was 1,333.33, Giving us a value of 0.2, 2, 2, 5V. So we have a 0.2, 225 volts. Then lastly, we have our gives for energy for the reaction. That's gonna be our negative 60.8 kg joules plus our gas constant 8.3 Times are temperature, which was 298 Kelvin Times our natural log of Q, which was 1,333.33. And that gives us a value of -42. kg joules. So we have negative 42.96 killed jules. And with that, we have answered the complete question overall, I hope this helped. And until next time.

Consider the reaction shown here occurring at 25°C. Cr(s) + Cd²⁺(aq) → Cr²⁺(aq) + Cd(s) Determine E°_cell, K, and ∆G°_rxn for the reaction and complete the table.
[Cd²⁺] [Cr²⁺] Q E_cell 𝚫G_rxn
1.00 1.00
1.00 1.00 × 10^-5
1.00 × 10^-5 1.00
4.18 × 10^-4 1.00

Verified Solution

Key Concepts

Electrochemical Cell

Nernst Equation

Gibbs Free Energy (ΔGrxn)

Consider the reaction shown here occurring at 25°C. Cr(s) + Cd2+(aq) → Cr2+(aq) + Cd(s) Determine E°cell, K, and ∆G°rxn for the reaction and complete the table. [Cd2+] [Cr2+] Q Ecell 𝚫Grxn 1.00 1.00 1.00 1.00 × 10-5 1.00 × 10-5 1.00 4.18 × 10-4 1.00

Verified Solution

Key Concepts

Electrochemical Cell

Nernst Equation

Gibbs Free Energy (ΔGrxn)

Consider the reaction shown here occurring at 25°C. Cr(s) + Cd²⁺(aq) → Cr²⁺(aq) + Cd(s) Determine E°_cell, K, and ∆G°_rxn for the reaction and complete the table.
[Cd²⁺] [Cr²⁺] Q E_cell 𝚫G_rxn
1.00 1.00
1.00 1.00 × 10^-5
1.00 × 10^-5 1.00
4.18 × 10^-4 1.00