Question

Find the equilibrium constant for a 3-electron transfer process, whose standard emf is `"0.59 V"` at `"298.15 K"`? `F = "96485.33 C/mol e"^(-)`

`a)` `10^15`
`b)` `10^20`
`c)` `10^25`
`d)` `10^30`

Answer 1

The standard EMF (electromotive "force") is `E_"cell"^@`. So, you should relate the following two equations:

`\mathbf(DeltaG = DeltaG^@ + RTlnQ)`

`\mathbf(DeltaG^@ = -nFE_"cell"^@)`

where:

• `DeltaG` is the Gibbs' free energy for the process.
• `DeltaG^@` is the Gibbs' free energy for the process at `mathbf(25^@ "C")` and `\mathbf("1 bar")` pressure.
• `R = "8.314472 J/mol"cdot"K"` is the universal gas constant.
• `T` is the temperature at which the process occurs, in `"K"`. For a standard thermodynamic process we assume it is `"298.15 K"`.
• `n` is the `"mol"`s of electrons transferred.
• `"F"` is Faraday's constant, which is approximately `"96485.33 C/mol e"^(-)`.
• `E_"cell"^@` is in `"V"`, and `"1 V"cdot"C" = "1 J"`.

Since we are finding `K_"eq"`, we know that `DeltaG = 0` and `Q = K` at equilibrium. Therefore:

`DeltaG^@ = -RTlnK_"eq" = -nFE_"cell"^@`

`=> color(blue)(K_"eq") = e^(nFE_"cell"^@"/"RT)`

`= e^(("3 mols e"^(-)"/1 mol atom"cdot"96485.33 C/mol e"^(-)cdot"0.59 V")"/"("8.314472 J/mol"cdot"K"cdot"298.15 K")`

`= e^68.89`

`~~ color(blue)(8.3xx10^(29))`

I found the actual answer to this elsewhere, and all they give is `10^30`, which is close enough, since the other given answer choices were `10^15`, `10^20`, and `10^25`.