Question

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?

Answer 1

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))`.