# For the reaction of "Au"^(3+) with "Cl"_2, if ["Cl"^(-)] and ["Au"^(3+)] were initially "0.10 M" and "0.25 M" respectively, while P_("Cl"_2) = "0.5 bar" initially, what is E_(cell)?

## Ecell=0.14V Q=2x106 Ecell=0.07785V

Jul 31, 2017

Well, what I would do is draw out the problem to see what you got...

The Nernst equation is:

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

• ${E}_{c e l l}$ is the cell potential at nonstandard $T$ and $P$. ""^@ signifies $\text{1 bar}$ and ${25}^{\circ} \text{C}$.
• $R$ and $T$ are known from the ideal gas law, and are in $\text{J/mol"cdot"K}$ and $\text{K}$, respectively.
• $n$ is the mols of electrons transferred in the overall reaction.
• $F = {\text{96485 C/mol e}}^{-}$ is the Faraday constant.
• $Q$ is the reaction quotient.

From the looks of it, you are supposed to calculate ${E}_{c e l l}$ at this nonstandard, nonequilibrium state.

First, write out the half-reactions. You need to know how to look up reduction reactions in a standard reduction potential table, which is mandatory at any chemistry exam with electrochemistry questions.

$\text{Au"^(3+)(aq) + 3e^(-) -> "Au} \left(s\right)$, $\text{ "E_(red)^@ = "1.50 V}$

${\text{Cl"_2(g) + 2e^(-) -> 2"Cl}}^{-} \left(a q\right)$, $\text{ "E_(red)^@ = "1.36 V}$

Since ${E}_{red}^{\circ} \left({\text{Au") > E_(red)^@("Cl}}_{2}\right)$, gold would rather be reduced than oxidized. Thus, the overall reaction at standard conditions is:

$2 \left(\text{Au"^(3+)(aq) + cancel(3e^(-)) -> "Au} \left(s\right)\right)$, $\text{ "E_(red)^@ = "1.50 V}$
$3 \left(2 {\text{Cl"^(-)(aq) -> "Cl}}_{2} \left(g\right) + \cancel{2 {e}^{-}}\right)$, $\text{ "E_(o x)^@ = -"1.36 V}$
$\text{-------------------------------------------}$
$2 \text{Au"^(3+)(aq) + 6"Cl"^(-)(aq) -> 3"Cl"_2(g) + 2"Au} \left(s\right)$

And its standard cell potential, i.e. ${E}_{c e l l}^{\circ}$ (and NOT ${E}_{c e l l}$ as you had stated!) is:

$\textcolor{b l u e}{\underline{{E}_{c e l l}^{\circ}}} = {E}_{red}^{\circ} + {E}_{o x}^{\circ}$

= "1.50 V" + (-"1.36 V") = ul(color(blue)(+"0.14 V"))

For this reaction, we have the reaction quotient, which is written just like $K$ but with current concentrations/partial pressures,

$Q = \left({\left({P}_{{\text{Cl"_2(g))//P^@)^3)/((["Au"^(3+)(aq)]//c^@)^2(["Cl}}^{-}} / {c}^{\circ}\right)}^{6}\right)$

where ${P}^{\circ} = \text{1 bar}$ and ${c}^{\circ} = \text{1 M}$ such that $Q$ is unitless. It must be, or else it cannot be operated on by $\ln$.

We can immediately obtain the reaction quotient with the given data:

color(blue)(ulQ) = ("0.5 bar"//"1 bar")^3/(("0.25 M"//"1 M")^2("0.10 M"//"1 M")^6)

$= \textcolor{b l u e}{\underline{2 \times {10}^{6}}}$

And that allows us to finally get the nonstandard cell potential. You didn't tell us the temperature, but I am guessing it is $\text{298.15 K}$.

color(blue)(ul(E_(cell))) = "0.14 V" - (("8.314472 V"cdot"C/mol"cdot"K")("298.15 K"))/(("6 mol e"^(-)/("1 mol atom"))("96485 C/mol e"^(-)))ln(2 xx 10^6)

$=$ $\text{0.07787 V}$

$\approx$ $\textcolor{b l u e}{\underline{\text{0.07785 V}}}$

And indeed it is at $\text{298.15 K}$. The answer wasn't exact because we were either using slightly different Faraday constants or universal gas constants, or your temperature was more like $\text{298 K}$.