Hi. Could someone walk me through how to identify the half reactions in the anode and cathode? I am aware that the left hand sides should contain Al and O2 but I don't know how to proceed.
The “aluminum–air fuel cell” is used as a reserve battery in remote locations. In this cell, aluminum reacts with the oxygen in air in basic solution. (a) Write the oxidation and reduction half- reactions for this cell.
textbook 6.73 [ENDORSED]
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Re: textbook 6.73 [ENDORSED]
The textbook discusses the aluminum-air fuel cell:
anode: 6OH-(aq) + 2Al(s) --> 2Al(OH)3 (aq) + 6e-
cathode: 3H2O(l) + (3/2)O2(g) + 6e ---> 6OH- (aq)
overall: 3H2O(l) + 2Al(s) + (3/2)O2(g) --> 2Al(OH)3 (aq)
Ecell = E cathode - E anode = (0.40V) - (-1.66V) = +2.06V
All standard reduction half-reactions will/are given.
As I communicated students are not expected to know by memory different batteries and fuel cells unless I cover them in class.
Hope this helps.
Since this is a question in the textbook students can look up the relevant information about specific batteries, fuel cells, etc.
anode: 6OH-(aq) + 2Al(s) --> 2Al(OH)3 (aq) + 6e-
cathode: 3H2O(l) + (3/2)O2(g) + 6e ---> 6OH- (aq)
overall: 3H2O(l) + 2Al(s) + (3/2)O2(g) --> 2Al(OH)3 (aq)
Ecell = E cathode - E anode = (0.40V) - (-1.66V) = +2.06V
All standard reduction half-reactions will/are given.
As I communicated students are not expected to know by memory different batteries and fuel cells unless I cover them in class.
Hope this helps.
Since this is a question in the textbook students can look up the relevant information about specific batteries, fuel cells, etc.
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