How to do this question for A and D (electrolysis half cells)?
1 Answer
Here's what I get.
Explanation:
Part a
The positive terminal of the battery connects to A. Thus,
-
A and C are the anodes (where oxidation occurs)
-
B and D are the cathodes (where reduction occurs)
Part b
1. Half-reaction at anode A
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#"C"# is an inert electrode. It will not be oxidized. -
#"Al"^"3+"# is already oxidized -
#"H"_2"O"# is the only remaining candidate.
In
The anode half-reaction is
#color(blue)("2H"_2"O" → "O"_2 + "4H"^"+" + "4e"^"-")#
2. Half-reaction at cathode B
The possibilities are:
#2"H"_2"O" + "2e"^"-" → "H"_2 + 2"OH"^"-" ; E^@ = "-0.82 V"# #"Al"^"3+" + 3"e"^"-" →"Al"; color(white)(mmmmll)E^@ = "-1.66 V"#
It is easier to reduce water than it is to reduce
The cathode half-reaction is
#color(blue)(2"H"_2"O" + "2e"^"-" → "H"_2 + 2"OH"^"-")#
3. Half-reaction at anode C
The possibilities are
-
#"Cu"^"2+" + 2"e"^"-" →"Cu"; color(white)(mmmll)E^@ = "+0.337"# -
#"O"_2 + "4H"^"+" + "4e"^"-" → "2H"_2"O"; E^@ = "+1.229 V"#
It is easier to oxidize copper than water, so the anode half-reaction is
#color(blue)("Cu" → "Cu"^"2+" + 2"e"^"-")#
4. Half-reaction at cathode D
The possibilities are:
#2"H"^"+" + "2e"^"-" → "H"_2; color(white)(mmmmll)E^@ = "0 V"# #2"H"_2"O" + "2e"^"-" → "H"_2 + 2"OH"^"-" ; E^@ = "-0.82 V"#
It is easier to reduce
The cathode half-reaction is
#color(blue)(2"H"^"+" + "2e"^"-" → "H"_2)#
