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

$\text{Cu (s)"\toCu^(2+)"(0.010 M)} + 2 {e}^{-}$ $M n {O}_{4}^{-} \text{(2.0 M)"+4H^+"(1.0 M)"+3e^(-)\toMnO_2"(s)"+2H_2O"(l)}$ I actually have everything... except ${E}_{c e l l}^{o}$, because I was not given voltages. Aug 2, 2018

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

So, I get $\text{1.40 V}$.

The values are:

$3 \left({\text{Cu"(s) -> "Cu}}^{2 +} \left(a q\right) + \cancel{2 {e}^{-}}\right)$, ${E}_{red}^{\circ} = \text{0.34 V}$
$\underline{2 \left(\text{MnO"_4^(-)(aq) + 4"H"^(+)(aq) + cancel(3e^(-)) -> "MnO"_2(s) + 2"H"_2"O} \left(l\right)\right)}$, ${E}_{red}^{\circ} = \text{1.67 V}$

$3 \text{Cu"(s) + 2"MnO"_4^(-)(aq) + 8"H"^(+)(aq) -> 3"Cu"^(2+)(aq) + 2"MnO"_2(s) + 4"H"_2"O} \left(l\right)$

Hence, we can calculate ${E}_{c e l l}^{\circ}$. Again, two ways I could do this.

${E}_{c e l l}^{\circ}$

$= \left\{\begin{matrix}{\overbrace{{E}_{red}^{\circ}}}^{\text{Reduction" + overbrace(E_(o x)^@)^"Oxidation" \\ underbrace(E_"cathode"^@)_"Reduction" - underbrace(E_"anode"^@)_"Reduction}}\end{matrix}\right.$

$= \left\{\begin{matrix}\text{1.67 V" + (-"0.34 V") \\ "1.67 V" - "0.34 V}\end{matrix}\right.$

$= + \text{1.33 V}$

As a result, one can then calculate ${E}_{c e l l}$ at these nonstandard concentrations from the Nernst equation.

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln Q$

where:

• $R$ and $T$ are known from the ideal gas law as the universal gas constant $\text{8.134 V"cdot"C/mol"cdot"K}$ and temperature in $\text{K}$.
• $n$ 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.
• $F = {\text{96485 C/mol e}}^{-}$ 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 ${c}^{\circ} = \text{1 M}$ to aqueous species):

$Q = \left({\left(\left[{\text{Cu"^(2+)]//c^@)^3)/((["MnO"_4^(-)]//c^@)^2(["H}}^{+}\right] / {c}^{\circ}\right)}^{8}\right)$

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

$= 2.5 \times {10}^{- 7}$

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

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))

$= {1.39}_{5}$ $\text{V}$

$\implies$ $\textcolor{b l u e}{\text{1.40 V}}$