Nickel and aluminum electrodes are used to build a galvanic cell. What is the theoretical cell potential assuming standard conditions? What is the shorthand notation for this cell?

1 Answer
Jul 4, 2016

Answer:

See below:

Explanation:

List the 1/2 equations in order least positive to most positive:

#sf(color(white)(xxxxxxxxxxxxxxxxxxxxxx)E^@("V"))#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#

#sf(Al_((aq))^(3+)" "+3e" "rightleftharpoons" "Al_((s))" "-1.662)#

#sf(Ni_((aq))^(2+)" "+2e" "rightleftharpoons" "Ni_((s))" "-0.25)#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)#

Note we use the #rightleftharpoons# symbol to show that the 1/2 cells can go in either direction depending on what they are coupled with.

The 1/2 cell with the most +ve #"E"^@# is the one which will take in the electrons.

From this we can see that the 2nd 1/2 cell will be driven left to right and the 1st 1/2 cell right to left in accordance with the arrows.

So the 2 half - reactions are:

#sf(Ni_((aq))^(2+)+2erarrNi_((s)))#

#sf(Al_((s))rarrAl_((aq))^(3+)+3e)#

To get the equation to balance we need to X the first 1/2 equation by 3 and the 2nd half equation by 2 then add to get the overall cell reaction:

#sf(3Ni_((aq))^(2+)+2Al_((s))rarr3Ni_((s))+2Al_((aq))^(3+))#

To calculate #sf(E_(cell)^(@))# you subtract the least positive electrode potential value from the most positive:

#sf(E_(cell)^(@)=-0.25-(-1.662)=+0.41" ""V")#

The shorthand notation is:

#sf(Al_((s))color(red)(I)color(white)(g)Al_((aq))^(3+)color(red)(\ll\)color(white)(4)Ni_((aq))^(2+)color(white)(4)color(red)(l)color(white)(5)Ni_((s)))#

Note the most oxidised form goes next to the salt bridge which is #sf(color(red)(ll))#.