Question #beb88
2 Answers
As per the symbolic presentation of cell the right hand half cell is cathode and the left hand half cell is anode.
And the EMF of the cell is
Here's what I got.
Explanation:
Just for the sake of your understanding, I'll add a bit of explanation on how this cell works.
Let's assume that you're not familiar with how the cell notation works for galvanic cells.
The problem provides you with two reduction half-reactions, mainly
#"M"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "M"_ ((s))" "E^@ = -"0.72 V"#
#"N"_ ((aq))^(3+) + 3"e"^(-) rightleftharpoons "N"_ ((s))" "E^@ = +"0.28 V"#
Now, those values given to you for the standard reduction potential,
A negative
A positive
Now, when you put these two half-cells together, the equilibrium with the more negative / less positive
In your case,
#"M"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "M"_ ((s))" "E^@ = -"0.72 V"#
#stackrel(color(blue)(larr))(color(white)(aacolor(red)("shift to the left")aaaa))#
This will be your oxidation half-reaction -- remember to reverse the sign of
#"M"_ ((s)) -> "M"_ ((aq))^(2+) + 2"e"^(-)" "E_"oxi"^@ = - (-"0.72 V") = +"0.72 V"#
Likewise, you will have
#"N"_ ((aq))^(3+) + 3"e"^(-) rightleftharpoons "N"_ ((s))" "E^@ = +"0.28 V"#
#stackrel(color(blue)(rarr))(color(white)(aacolor(red)("shift to the right")aaaa))#
This will be your reduction half-reaction
#"N"_ ((aq))^(3+) + 3"e"^(-) -> "N"_ ((s))" "E_"red"^@ = +"0.28 V"#
Balance and add the two half-reactions to get
#{(color(white)(aaaaaa)"M"_ ((s)) -> "M"_ ((aq))^(2+) + 2"e"^(-)" "| xx 3), ("N"_ ((aq))^(3+) + 3"e"^(-) -> "N"_ ((s))" "color(white)(aaaaaaa)| xx 2) :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
#3"M"_ ((s)) + 2"N"_ ((aq))^(3+) + color(red)(cancel(color(black)(6"e"^(-)))) -> 3"M"_ ((aq))^(2+) + color(red)(cancel(color(black)(6"e"^(-)))) + 2"N"_ ((s))#
You will thus have
#3"M"_ ((s)) + 2"N"_ ((aq))^(3+) -> 3"M"_ ((aq))^(2+) + 2"N"_ ((s))#
Since you can add the two half-reactions, you can do the same with the
You will thus have
#E_"cell"^@ = E_"oxi"^@ + E_"red"^@#
This gets you
#E_"cell"^@ = +"0.72 V" + "0.28 V" = "1.00 V"#
A quick word about galvanic cell notation.
As you can see, the species that is being oxidized is always added first. In your case,
The species that is being reduced is always added last. In your case,
#overbrace("M"_ ((s)) | "M"_ ((aq))^(2+))^(color(blue)("what is being oxidized"))" " || " "overbrace("N"_ ((aq))^(3+) | "N"_ ((s)))^(color(darkgreen)("what is being reduced"))#
So next time you are given a cell notation, reverse