Question

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.
`Pt` (s) `"| "Cr^(2+)` (0.30 M), `Cr^(3+)` (2.0 M) `"|| " Co` (0.20 M) `"| "Co` (s)

"For the following galvanic cell run at 25°C, if the equilibrium constant `K` is `2.79xx10^7`, calculate `\DeltaG` at these conditions?"

Answer 1

`DeltaG = -"29.10 kJ/mol"`

---

This is just a way to get you to work more with the shift in `DeltaG` from standard conditions:

`DeltaG = DeltaG^@ + RTlnQ`

where `@` indicates standard conditions, like `"1 M"` concentrations, `"1 atm"` partial pressures, etc. for a given temperature.

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

`cancel(DeltaG)^(0) = DeltaG^@ + RTlncancel(Q)^(K)`

Thus, assuming this is `K_c`,

`DeltaG^@ = -RTlnK_c`

`= -8.314 cancel"J""/mol"cdotcancel"K" xx "1 kJ"/(1000 cancel"J") cdot 298.15 cancel"K" cdot ln(2.79 xx 10^7)`

`= -"42.50 kJ/mol"`

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

`2("Cr"^(2+)(aq) -> "Cr"^(3+)(aq) + e^(-))`, `E_(red)^@ = -"0.41 V"`
`ul("Co"^(2+)(aq) + 2e^(-) -> "Co"(s))`, `E_(red)^@ = -"0.277 V"`
`2"Cr"^(2+)(aq) + "Co"^(2+)(aq) -> "Co"(s) + 2"Cr"^(3+)(aq)`

We don't need `E_(cell)^@`, but we did need to figure out what direction was spontaneous. Since `E_(red)^@` for `"Cr"^(2+ ->3+)` was more negative, that oxidation is spontaneous and we chose correctly.

(Also, `"Co"^(2+)` is more likely than `"Co"^+`, so that has the `"0.20 M"` concentration...)

`Q = (["Cr"^(3+)]^2)/(["Cr"^(2+)]^2["Co"^(2+)])`

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

`= 222.22`

Finally,

`color(blue)(DeltaG) = overbrace(-"42.50 kJ/mol")^(DeltaG^@) + overbrace("0.008314 kJ/mol"cdotcancel"K" cdot 298.15 cancel"K" cdot ln(222.22))^(RTlnQ)`

`=` `color(blue)(-"29.10 kJ/mol")`

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