Question #beb88

2 Answers
Aug 9, 2016

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

#E_"cell"=E_"cathode"-E_"anode"#
#=E_(N^"3+""/"N)- E_(M^"2+""/"M )=(0.28-(-0.72))V=1V#

Aug 9, 2016

Answer:

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, #E^@#, tell you where the position of the reduction equilibrium stands when compared with a reference electrode, usually a hydrogen one.

A negative #E^@# value means that the equilibrium will lie to the left, i.e. the chemical species loses electrons more readily than hydrogen.

A positive #E^@# means that the equilibrium will lie to the right, i.e. the chemical species loses electrons less readily than hydrogen.

Now, when you put these two half-cells together, the equilibrium with the more negative / less positive #E^@# value will shift to the left and the equilibrium with the less negative / more positive #E^@# value will shift to the right.

In your case, #E^@# is negative for the first equilibrium and positive for the second one, which means that you'll have

#"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 #E^@# !!!

#"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 #E_"oxi"^@# and #E_"red"^@# values -- keep in mind that multiplying the half-reaction by a given coefficient does not change the value of #E^@# !!!

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.

https://chemistry59.wikispaces.com/Electrochemistry

As you can see, the species that is being oxidized is always added first. In your case, #"M"_ ((s))# is being oxidized to #"M"_ ((aq))^(2+)#.

The species that is being reduced is always added last. In your case, #"N"_ ((aq))^(3+)# is being reduced to #"N"_ ((s))#. This is why your cell notation looks like this

#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 #E^@# value for the species that is being oxidized, keep the #E^@# value for the species that is beign reduced, and just add the resulting values to get #E_"cell"^@#.