Arrange the "E"^(@) values in increasing order:
" " "E"^(@)("V")
color(red)stackrel(leftarrow)(color(white)(xxxxxxxxxxxxx)
Fe_((aq))^(3+)+erightleftharpoonsFe_((aq))^(2+)" "+0.77
MnO_(4(aq))^(-)+8H_((aq))^(+)+5erightleftharpoonsMn_((aq))^(2+)+4H_2O_((l))" "+1.51
color(blue)stackrel(rightarrow)(color(white)(xxxxxxxxxxxxxxxxxxxxxx)
The Mn(VII) 1/2 cell has the more +ve value so the reaction will be driven in the direction indicated by the arrows
To find "E"_(cell)^(@) you subtract the least positive value from the most +ve:
"E"_(cell)^(@)=1.51-0.77=+0.74"V".
The Nernst Equation at 25^(@)"C" can be written:
E_(cell)=E_(cell)^(@)-0.05916/(n)log"Q"
"Q" is the reaction quotient
n is the number of moles of electrons transferred, which in this case =5
As the reaction proceeds, the potential difference between the two 1/2 cells falls.
When it reaches zero the reaction has reached equilibrium so now we can write:
0=E_(cell)^(@)-0.05916/(5)log"K"
Note that "Q" has been replaced by "K", the equilibrium constant.
Putting in the value for E_(cell)^(@) and rearranging rArr
log"K"=(0.74xx5)/(0.05916)=62.5
:."K"=3.48xx10^(62)
This is a number so large that we can say that the reaction has gone to completion.
