the two half-reactions that are used in the direct methanol fuel cell are shown below.
I. 2 CH3OH(aq)+2H2O(l)=>2CO2(g)+12H+(aq)+12e-
II. 3O2(g)+12H+(aq) +12e-=>6H2O(l)
the electrode at which half-reaction I occurs is the ______ at which______ is oxidized, and half-reaction II occurs at the ______ at which ________ is reduced.
need help with the blanks!
See your post above.
To determine the blanks in the statement, we need to understand the concepts of oxidation and reduction in the context of the given half-reactions.
In a redox (reduction-oxidation) reaction, one species is oxidized (loses electrons), while another species is reduced (gains electrons).
Now let's analyze the given half-reactions:
I. 2 CH3OH(aq) + 2H2O(l) => 2CO2(g) + 12H+(aq) + 12e-
II. 3O2(g) + 12H+(aq) + 12e- => 6H2O(l)
In half-reaction I, methanol (CH3OH) is being oxidized, as it loses electrons and its oxidation state increases from -2 to +4 in the reaction. Therefore, half-reaction I occurs at the electrode where oxidation takes place.
Similarly, in half-reaction II, oxygen (O2) is being reduced, as it gains electrons and its oxidation state decreases from 0 to -2 in the reaction. Therefore, half-reaction II occurs at the electrode where reduction takes place.
Now we can fill in the blanks:
The electrode at which half-reaction I occurs is the **anode** at which **methanol is oxidized**, and half-reaction II occurs at the **cathode** at which **oxygen is reduced**.
In a nutshell, the half-reaction I occurs at the anode where methanol gets oxidized, and the half-reaction II occurs at the cathode where oxygen gets reduced in a direct methanol fuel cell.