Question

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)"`

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

Answer 1

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

So, I get `"1.40 V"`.

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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")`