Calculate #E_(cell)# for a battery based in the following two half-reactions and conditions?

#"Cu (s)"\toCu^(2+)"(0.010 M)"+2e^-#
#MnO_4^(-)"(2.0 M)"+4H^+"(1.0 M)"+3e^(-)\toMnO_2"(s)"+2H_2O"(l)"#


I actually have everything... except #E_(cell)^o#, because I was not given voltages.

webcam

1 Answer
Aug 2, 2018

You should be given them, or you can look them up. I found them to be:#""^([1])##""^([2])#

So, I get #"1.40 V"#.


The values are:

#3("Cu"(s) -> "Cu"^(2+)(aq) + cancel(2e^(-)))#, #E_(red)^@ = "0.34 V"#
#ul(2("MnO"_4^(-)(aq) + 4"H"^(+)(aq) + cancel(3e^(-)) -> "MnO"_2(s) + 2"H"_2"O"(l)))#, #E_(red)^@ = "1.67 V"#

#3"Cu"(s) + 2"MnO"_4^(-)(aq) + 8"H"^(+)(aq) -> 3"Cu"^(2+)(aq) + 2"MnO"_2(s) + 4"H"_2"O"(l)#

Hence, we can calculate #E_(cell)^@#. Again, two ways I could do this.

#E_(cell)^@#

#= {(overbrace(E_(red)^@)^"Reduction" + overbrace(E_(o x)^@)^"Oxidation"),(underbrace(E_"cathode"^@)_"Reduction" - underbrace(E_"anode"^@)_"Reduction"):}#

#= {("1.67 V" + (-"0.34 V")),("1.67 V" - "0.34 V"):}#

#= +"1.33 V"#

As a result, one can then calculate #E_(cell)# at these nonstandard concentrations from the Nernst equation.

#E_(cell) = E_(cell)^@ - (RT)/(nF)lnQ#

where:

  • #R# and #T# are known from the ideal gas law as the universal gas constant #"8.134 V"cdot"C/mol"cdot"K"# and temperature in #"K"#.
  • #n# is the total mols of electrons per mol of the atoms involved. Just take its value to be the number of electrons cancelled out from the balanced reaction.
  • #F = "96485 C/mol e"^(-)# is Faraday's constant.
  • #Q# is the reaction quotient, i.e. the not-yet equilibrium constant.

#Q# is known, remembering that pure liquids and solids are given "concentrations" of #1#, and that we assign a standard concentration of #c^@ = "1 M"# to aqueous species):

#Q = ((["Cu"^(2+)]//c^@)^3)/((["MnO"_4^(-)]//c^@)^2(["H"^(+)]//c^@)^8)#

#= ("0.010 M"//"1 M")^3/(("2.0 M"//"1 M")^2("1.0 M"//"1 M")^8)#

#= 2.5 xx 10^(-7)#

From here, assuming you mean #T = "298.15 K"#, and knowing that #n = "6 mol e"^(-)"/mol atoms"#,

#color(blue)(E_(cell)) = "1.33 V" - ("8.314 V"cdotcancel"C"//cancel"mol"cdotcancel"K" cdot 298.15 cancel"K")/(((6 cancel("mol e"^(-)))/cancel"1 mol atoms") cdot 96485 cancel"C"//cancel("mol e"^(-)))ln(2.5 xx 10^(-7))#

#= "1.33 V" - ("0.0257 V")/(6)ln(2.5 xx 10^(-7))#

#= "1.33 V" - ("0.0592 V")/(6)log(2.5 xx 10^(-7))#

#= 1.39_5# #"V"#

#=># #color(blue)("1.40 V")#