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 worked this for you at your earlier post.
To write a net equation for the reaction in a methane fuel cell, we need to combine the two half-reactions provided. Here's how you can do it:
Step 1: Balance the number of electrons in each half-reaction.
In the first half-reaction, 8 electrons are involved, while in the second half-reaction, 4 electrons are involved. To balance the electron numbers, we multiply the first half-reaction by 2:
2CO32-(aq) + 14H2O(g) + 16e- -----> 2CH4(g) + 20OH-(aq)
Step 2: Adjust the coefficients to make sure the number of atoms is balanced.
The number of water molecules on the reactant side differs between the two half-reactions. To balance this, we multiply the second half-reaction by 7:
2CO32-(aq) + 14H2O(g) + 16e- -----> 2CH4(g) + 20OH-(aq)
7O2(g) + 14H2O(g) + 28e- -----> 28OH-(aq)
Now, we combine the two half-reactions to get the net equation:
2CO32-(aq) + 14H2O(g) + 16e- + 7O2(g) + 14H2O(g) + 28e- -----> 2CH4(g) + 20OH-(aq) + 28OH-(aq)
Simplifying it further:
2CO32-(aq) + 28H2O(g) + 44e- + 7O2(g) -----> 2CH4(g) + 48OH-(aq)
This is the net equation for the reaction in the methane fuel cell.
To calculate the potential for the methane fuel cell, we need to find the net potential (Ecell) by subtracting the reduction potential of the anode (Ero_anode) from the reduction potential of the cathode (Ero_cathode).
Ecell = Ero_cathode - Ero_anode
Given:
Ero_anode = 0.17 V
Ero_cathode = 0.40 V
Substituting the values into the formula:
Ecell = 0.40 V - 0.17 V
Ecell = 0.23 V
Therefore, the potential for the methane fuel cell is 0.23 V.