In a methane fuel cell, the chemical energy of the methane is converted into electrical energy instead of heat that would flow during the combustion of methane. Using the half-reactions and reduction potentials given below
a) Write a net equation for the reaction.
b) Calculate the potential for the methane fuel cell.
CO32-(aq) + 7H2O(g) + 8e- -----> CH4(g) + 10OH-(aq) Ero = +0.17 V
O2(g) + 2H2O(g) + 4e- ------> 4OH-(aq) Ero = +0.40 V
I don't know what Ero stands for but these are reduction potentials you have written. Multiply equation 2 by 2, reverse it, add to equn 1, and eliminate ions common to both sides (for example you can eliminate some of the OH and some of the H2O) and that will be the net ionic equation.
For b part, reverse equation 2 (and change the sign of Eo), add to equation 1 and that will be the cell potential.
As you saw in Part B, the vapor above the cyclohexane-acetone solution is composed of both cyclohexane vapor and acetone vapor. What mole fraction of the vapor above the solution, , is cyclohexane
To write the net equation for the reaction in a methane fuel cell, we need to combine the half-reactions and balance the equations. The first step is to make sure that the number of electrons transferred is the same in both half-reactions.
In the given half-reactions:
1) CO32-(aq) + 7H2O(g) + 8e- -----> CH4(g) + 10OH-(aq)
2) O2(g) + 2H2O(g) + 4e- -----> 4OH-(aq)
We can see that the second half-reaction has 2 electrons involved, while the first half-reaction has 8 electrons involved. To balance the electron transfer, we will multiply the second half-reaction by 4, so that both reactions involve 8 electrons:
1) CO32-(aq) + 7H2O(g) + 8e- -----> CH4(g) + 10OH-(aq)
2) 4O2(g) + 8H2O(g) + 16e- -----> 16OH-(aq)
Now, we can add the two reactions together:
1) CO32-(aq) + 7H2O(g) + 8e- -----> CH4(g) + 10OH-(aq)
2) 4O2(g) + 8H2O(g) + 16e- -----> 16OH-(aq)
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Net Equation: 4CO32-(aq) + 7H2O(g) + 4O2(g) -----> CH4(g) + 4CO2(g) + 20OH-(aq)
Now, to calculate the potential for the methane fuel cell, we can use the reduction potentials (Ero) given for each half-reaction.
The overall potential (Ecell) is given by the formula:
Ecell = Ero(cathode) - Ero(anode)
In this case, the reduction potential for the methane half-reaction (cathode) is +0.17 V, and the reduction potential for the oxygen half-reaction (anode) is +0.40 V.
So, the potential for the methane fuel cell can be calculated as:
Ecell = +0.17 V - (+0.40 V)
Ecell = +0.17 V - 0.40 V
Ecell = -0.23 V
Therefore, the potential for the methane fuel cell is -0.23 V.