Given the half reactions Au+3 +3e- -> Au and Sn+4 +2e- -> Sn+2 in a galvanic cell, what is the voltage if [Au+3]=1.45M [Sn+4]=.50M and [Sn+2]=.87M?

The first thing to do is to calculate the emf of the cell under standard conditions.

We can do this using standard electrode potentials (`sf(E^@)`):

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

`" " "E"^@("V")`

`stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)`

`sf(Sn^(4+)" "+" "2e" "rightleftharpoons" "Sn^(2+)" "+0.15)`

`sf(Au^(3+)" "+" "3e" "rightleftharpoons" "Au" "+1.52)`

`stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)`

The more +ve half - cell will take in the electrons so the reactions in each half - cell will proceed in the direction shown by the arrows.

This gives the overall cell reaction:

`sf(2Au^(3+)+3Sn^(2+)rarr2Au+3Sn^(4+))`

To find `sf(E_(cell)^@)` subtract the least +ve electrode potential from the most positive:

`sf(E_(cell)^@=+1.52-(+0.15)=1.37color(white)(x)V)`

Since we are not under standard conditions we now need to use The Nernst Equation.

A useful form of this at `sf(25^@C)` is:

`sf(E_(cell)=E_(cell)^@-(0.05916)/(z)logQ)`

`sf(z)` is the number of moles of electrons transferred which, in this case = 6.

`sf(Q)` is the reaction quotient and is given by:

`sf(Q=([Sn^(4+)]^(3))/([Au^(3+)]^(2)[Sn^(2+)]^(3))`

`:.``sf(Q=((0.5)^(2))/((1.45)^(2)xx(0.87)^(3))=0.0902color(white)(x)"mol"^(-2).l^(2))`

`:.``sf(E_(cell)=1.37-(0.05916)/(6)log[0.0902])`

`sf(E_(cell)=1.37+0.103=+1.38color(white)(x)V)`