# Galvanic cell run at 25° C: calculate \DeltaG when K=2.79xx10^7?

## I really don't remember if I've posted this before, but. $P t$ (s) $\text{| } C {r}^{2 +}$ (0.30 M), $C {r}^{3 +}$ (2.0 M) $\text{|| } C o$ (0.20 M) $\text{| } C o$ (s) "For the following galvanic cell run at 25°C, if the equilibrium constant $K$ is $2.79 \times {10}^{7}$, calculate $\setminus \Delta G$ at these conditions?"

Aug 1, 2018

$\Delta G = - \text{29.10 kJ/mol}$

This is just a way to get you to work more with the shift in $\Delta G$ from standard conditions:

$\Delta G = \Delta {G}^{\circ} + R T \ln Q$

where $\circ$ indicates standard conditions, like $\text{1 M}$ concentrations, $\text{1 atm}$ partial pressures, etc. for a given temperature.

If we were to have been at equilibrium (which we're not), then

${\cancel{\Delta G}}^{0} = \Delta {G}^{\circ} + R T \ln {\cancel{Q}}^{K}$

Thus, assuming this is ${K}_{c}$,

$\Delta {G}^{\circ} = - R T \ln {K}_{c}$

$= - 8.314 \cancel{\text{J""/mol"cdotcancel"K" xx "1 kJ"/(1000 cancel"J") cdot 298.15 cancel"K}} \cdot \ln \left(2.79 \times {10}^{7}\right)$

$= - \text{42.50 kJ/mol}$

Next, the reaction quotient $Q$ is for the spontaneous reaction:

$2 \left({\text{Cr"^(2+)(aq) -> "Cr}}^{3 +} \left(a q\right) + {e}^{-}\right)$, ${E}_{red}^{\circ} = - \text{0.41 V}$
$\underline{\text{Co"^(2+)(aq) + 2e^(-) -> "Co} \left(s\right)}$, ${E}_{red}^{\circ} = - \text{0.277 V}$
$2 {\text{Cr"^(2+)(aq) + "Co"^(2+)(aq) -> "Co"(s) + 2"Cr}}^{3 +} \left(a q\right)$

We don't need ${E}_{c e l l}^{\circ}$, but we did need to figure out what direction was spontaneous. Since ${E}_{red}^{\circ}$ for ${\text{Cr}}^{2 + \to 3 +}$ was more negative, that oxidation is spontaneous and we chose correctly.

(Also, ${\text{Co}}^{2 +}$ is more likely than ${\text{Co}}^{+}$, so that has the $\text{0.20 M}$ concentration...)

$Q = \left(\left[{\text{Cr"^(3+)]^2)/(["Cr"^(2+)]^2["Co}}^{2 +}\right]\right)$

= ("2.0 M"//"1 M")^2/(("0.30 M"//"1 M")^2("0.20 M"//"1 M"))

$= 222.22$

Finally,

$\textcolor{b l u e}{\Delta G} = {\overbrace{- \text{42.50 kJ/mol")^(DeltaG^@) + overbrace("0.008314 kJ/mol"cdotcancel"K" cdot 298.15 cancel"K} \cdot \ln \left(222.22\right)}}^{R T \ln Q}$

$=$ $\textcolor{b l u e}{- \text{29.10 kJ/mol}}$

So even though this reaction is less spontaneous with these concentrations at this temperature, it is still spontaneous.