Question #f7401

1 Answer
May 22, 2016

Answer:

#"K"~=10^(62)# which can be regarded as infinitely large.

Explanation:

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Note that #"Q"# has been replaced by #"K"#, the equilibrium constant.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