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?

1 Answer
Jan 12, 2017

Answer:

#sf(E_(cell)=+1.38color(white)(x)V)#

Explanation:

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)#