Consider the following reaction: NaI (aq) + 3 HOCl (aq) --------> NaIO3 (aq) + 3

Question

Consider the following reaction:

NaI (aq) + 3 HOCl (aq) --------> NaIO3 (aq) + 3 HCl (aq)

     E degrees cell = 0.40 V

a.) calculate the Keq for this reaction

b.) calculate the delta G degrees for this reaction

c.) Is this reaction spontaneous as written?

d.) Calculate the Delta G when [NaI]= 0.1 M, [HOCl]= 0.5 M, [NaIO3]= 1.5 M, [HCl]= 1.2 M

e.) Is this reaction spontaneous under these conditions?

Please show all work, I am studying for an exam and this is a sample question that may appear on there and I would really like to know how to derive each answer. Thanks!!

Explanation / Answer

Half-rxns:

Oxidation (occurs at anode): I^-(aq) + 3H2O(l) = IO3^-(aq) + 6H^+(aq) + 6e^-

Reduction (occurs at cathode): HOCl(aq) + 2H^+(aq) + 2e^- = HCl(aq) + H2O(l)

Balance the electrons in both half-rxns so they cancel when both half-rxns are combined:

The first half-rxn remains as is and the second on is multiplied by 3.

3(HOCl(aq) + 2H^+(aq) + 2e^- = HCl(aq) + H2O(l))

becomes

3HOCl(aq) + 6H^+(aq) + 6e^- = 3HCl(aq) + 3H2O(l)

Combine both half-rxns to give a balanced cell reaction:

I^-(aq) + 3H2O(l) + 3HOCl(aq) + 6H^+(aq) + 6e^- = IO3^-(aq) + 6H^+(aq) + 3HCl(aq) + 3H2O(l) + 6e^- ... electrons MUST cancel

I^-(aq) + 3HOCl(aq) + IO3^-(aq) + 3HCl(aq) ... this is the balanced cell reaction

a)

log Keq = nE^o(cell) / 0.0592

log Keq = (6) (0.40) / 0.0592

log Keq = 40.5

Keq = 10^40.5 = 3.16 x 10^40

b)

delta G^o(rxn) = -RTlnKeq

delta G^o(rxn) = -(8.314 J/mol*K) (298 K) (ln 3.16 x 10^40)

delta G^o(rxn) = -2.31 x 10^5 J/mol = -231 kJ/mol

c)

Since delta G^o(rxn) < 0, the reaction is spontaneous.

d)

Keq = [NaIO3] [HCl]^3 / [NaI] [HOCl]^3

Keq = (1.5) (1.2)^3 / (0.1) (0.5)^3 = 207.36

delta G^o(rxn) = -RTlnKeq

delta G^o(rxn) = -(8.314 J/mol*K) (298 K) (207.36)

delta G^o(rxn) = -5.14 x 10^5 J/mol = -514 kJ/mol

e)

Since delta G^o(rxn) < 0, the reaction is spontaneous.

Hope this helps! :)

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