Question

For the overall reaction given below at `25^@"C"`, if a saturated solution has formed, then if `["I"^-]_i = "0.500 M"` and the concentration has increased of the products by `1.33 xx 10^(-3) "M"` to reach equilibrium, what is `["I"_3^(-)]` at equilibrium?

`"I"_2(aq) + "I"^(-)(aq) -> "I"_3^(-)(aq)`

`E_(red)^@ = "0.5355 V"` for reduction of `"I"_2(aq)` to `"I"^(-)(aq)`.
`E_(red)^@ = "0.5365 V"` for reduction of `"I"_3^(-)(aq)` to `"I"^(-)(aq)`.

Answer 1

I got `["I"_3^(-)] = 6.17 xx 10^(-4) "M"`.

---

Since we're looking at a saturated solution, we're in an equilibrium condition such that

`DeltaG = -nFE_(cell) = 0`

but

`DeltaG^@ = -nFE_(cell)^@ ne 0`,

where:

• `DeltaG` is the change in Gibbs' free energy for the process.
• `n` is the mols of electrons per mols of reactants.
• `F = "96485 C/mol e"^(-)` is Faraday's constant.
• `E_(cell)^@` is the cell potential at standard conditions (`25^@ "C"` and `"1 atm"`).

The first `E_(red)^@` is for the reaction

`"I"_2(aq) + 2e^(-) rightleftharpoons 2"I"^(-)(aq)`

and the second `E_(red)^@` is for the reaction

`"I"_3^(-)(aq) + 2e^(-) rightleftharpoons 3"I"^(-)(aq)`

We can verify that here on page 2-3. For the overall reaction, we then have:

`"I"_2(aq) + cancel(2e^(-)) rightleftharpoons cancel(2"I"^(-)(aq)),` `E_(red)^@ = "0.5355 V"`

`ul(cancel3"I"^(-)(aq) rightleftharpoons "I"_3^(-)(aq) + cancel(2e^(-))),` `E_(o x)^@ = -"0.5365 V"`
`"I"_2(aq) + "I"^(-)(aq) rightleftharpoons "I"_3^(-)(aq)`

This adds up as (where `E_(o x)^@ = -E_(red)^@`):

`E_(cell)^@ = E_(red)^@ + E_(o x)^@`

`= "0.5355 V" - "0.5365 V"`

`= -"0.0010 V"`

So apparently we're looking at the nonspontaneous direction at this temperature. We'd just find the upside-down equilibrium constant.

Anyways, it doesn't matter since we're just calculating an equilibrium concentration of this reaction.

`DeltaG_(rxn)^@ = -nFE_(cell)^@`

`= -("2 mol e"^(-)"/mol I")("96485 C/mol e"^(-))(-"0.0010 V")`

`= +"192.97 J/mol"`

From

`cancel(DeltaG_(rxn))^(0) = DeltaG_(rxn)^@ + RTlncancel(Q)^(K)`,

we know that `DeltaG_(rxn)^@ = -RTlnK`, and we find the equilibrium constant to be:

`K_c = e^(-DeltaG_(rxn)^@//RT)`

`= e^(-"192.97 J/mol"//("8.314472 J/mol"cdot"K" cdot "298.15 K"))`

`= 0.9251`

This mass action expression would then be:

`K_c = 0.9251 = (["I"_3^(-)]_(eq))/(["I"_2]_(eq)["I"^(-)]_(eq))`

`= (["I"_3^(-)]_(eq))/(1.33xx10^(-3) "M" cdot ("0.500 M" + 1.33xx10^(-3) "M"))`

I assumed you knew how to do ICE tables at this point.

We started with `"0 M"` `"I"_2(aq)`, so if we saturate the solution, then `Delta["I"_2(aq)] = 1.33 xx 10^(-3) "M"`, and with coefficients of `1` all the way across, `["I"^(-)]_(eq) = ["I"^(-)]_i + Delta["I"_2(aq)]`.

Therefore, at equilibrium,

`color(blue)(["I"_3^(-)]_(eq)) = 0.9251 cancel("M"^(-1)) cdot 1.33xx10^(-3) cancel"M" cdot (0.500 cancel"M" + 1.33xx10^(-3) cancel"M")`

`= color(blue)(6.17 xx 10^(-4) "M")`