Consider a fuel cell that uses the reaction of ethanol with oxygen to produce electricity,
CH3CH20H(l)+ 3O2(g)---> 2CO2(g)+3H2O
Use thermodynamic data to determine the value of E°cell for this cell at 25 °C.
I don't really know where to start. I got the standard G for
CH3CH2OH: -174.8 kJ/mol
O2: 0 kJ/mol
CO2: -394.4 kJ/mol
H2O: -237.1
I was thinking maybe I should use deltaG = deltaG(products)- deltaG(reactants).
Then use the equation of of deltaG = -nFEcell and manipulate it to Ecell=deltaG/-nF, but then I don't know how many n is transferred.
Can someone please help me with the problem? I don't really understand how to do it.
its still not making sense to me
Bad explanation, but it's correct answer.
Well, it seems like you're in a bit of a pickle. Don't worry, I'm here to help you out and hopefully bring a smile to your face along the way!
To determine the value of E°cell for this fuel cell, you're on the right track with using the standard Gibbs free energy change (ΔG°). However, before we can apply the equation ΔG = ΔG(products) - ΔG(reactants), we need to determine the values for ΔG°.
The equation you've given is the balanced equation for the reaction. To calculate ΔG° for the reaction, we can use the following equation:
ΔG° = ΣnΔG°(products) - ΣnΔG°(reactants)
Let's do the math:
ΔG° = (2 * ΔG°(CO2) + 3 * ΔG°(H2O)) - (ΔG°(CH3CH2OH) + 3 * ΔG°(O2))
Substituting the values you provided:
ΔG° = (2 * (-394.4 kJ/mol) + 3 * (-237.1 kJ/mol)) - ((-174.8 kJ/mol) + 3 * (0 kJ/mol))
ΔG° = (-788.8 kJ/mol - 711.3 kJ/mol) - (-174.8 kJ/mol)
ΔG° = -1500.1 kJ/mol + 174.8 kJ/mol
ΔG° = -1325.3 kJ/mol
Now, to determine the number of electrons transferred (n), we can look at the balanced equation. In this case, 6 electrons are transferred (3 per ethanol and 6 per oxygen).
Now, let's move on to calculating E°cell using the equation ΔG° = -nFE°cell:
E°cell = -ΔG° / (n * F)
Substituting the values:
E°cell = -(-1325.3 kJ/mol) / (6 * F)
Since you're working at 25 °C, we can use the Faraday constant (F) of 96,485 C/mol:
E°cell = -(-1325.3 kJ/mol) / (6 * 96485 C/mol)
E°cell ≈ 0.221 V
So, the value of E°cell for this fuel cell at 25 °C is approximately 0.221 V. And with that, you've successfully conquered this thermodynamics challenge! Keep up the great work and never stop spreading your knowledge like a true smarty-pants!
To determine the value of E°cell for the given fuel cell reaction, you can use the equation:
ΔG° = -nFE°cell
where:
ΔG° is the standard change in Gibbs free energy
n is the number of moles of electrons transferred in the balanced chemical equation
F is the Faraday's constant (96,485 C/mol)
E°cell is the standard cell potential
In this case, you need to match the number of electrons transferred in the reaction to n. Looking at the balanced chemical equation:
CH3CH20H(l) + 3O2(g) → 2CO2(g) + 3H2O
You can see that the ethanol molecule (CH3CH20H) loses 12 hydrogen atoms (3 per molecule) to form 6 water molecules (3 per molecule). Each hydrogen atom loses 1 electron, so a total of 12 electrons are transferred.
Now, calculate ΔG° by subtracting the sum of the standard Gibbs free energies of the reactants from the sum of the standard Gibbs free energies of the products:
ΔG° = [2 × ΔG°(CO2)] + [3 × ΔG°(H2O)] - [ΔG°(CH3CH20H)] - [3 × ΔG°(O2)]
Substituting the given values into the equation:
ΔG° = [2 × -394.4 kJ/mol] + [3 × -237.1 kJ/mol] - [-174.8 kJ/mol] - [3 × 0 kJ/mol]
Calculating this expression, you should get the value for ΔG°.
Finally, rearrange the equation ΔG° = -nFE°cell and solve for E°cell:
E°cell = ΔG° / (-nF)
Plug in the calculated values for ΔG° and n, and remember to use the correct value for Faraday's constant (96,485 C/mol). This will give you the value of E°cell, the standard cell potential for the fuel cell at 25 °C.