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Note that #"Q"# has been replaced by #"K"#, the equilibrium constant.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.
