Calculate E_(cell)Ecell for a battery based in the following two half-reactions and conditions?
"Cu (s)"\toCu^(2+)"(0.010 M)"+2e^-Cu (s)→Cu2+(0.010 M)+2e−
MnO_4^(-)"(2.0 M)"+4H^+"(1.0 M)"+3e^(-)\toMnO_2"(s)"+2H_2O"(l)"MnO−4(2.0 M)+4H+(1.0 M)+3e−→MnO2(s)+2H2O(l)
I actually have everything... except E_(cell)^oEocell , because I was not given voltages.
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I actually have everything... except
1 Answer
You should be given them, or you can look them up. I found them to be:
So, I get
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)^@
= {(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) = E_(cell)^@ - (RT)/(nF)lnQ where:
R andT 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 = ((["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
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")
