Consider the following two half-reactions and their standard reduction potential

Question

Consider the following two half-reactions and their standard reduction potentials. Answer the three questions below. O_3(g) + H_2O(l) + 2e^- rightarrow O_2(g) + 2OH^-(aq) E degree = 1.246 V ClO^-_3 (aq) + 3H_2O(l) + 6e^- rightarrow Cl^-(aq) + 6 OH^- (aq) E degree = 0.622 V (a). Calculate E degree for the spontaneous redox reaction that occurs when these two half-reactions are coupled. Number (b). Calculate the value of Delta G degree for the reaction. Number kJ (c). Determine the equilibrium constant for the reaction. Number.

Explanation / Answer

The reactions are O3(g)+ H2O(l) +2e- --àO2(g)+2OH-(aq)   Eo= 1.246V, (1)

Multiplying the equation by 3 gives

3O3(g)+ 3H2O(l) +6e-----à3O2+6OH--, Eo= 1.246V (1)

ClO3-(aq)+3H2O(l) +6e- --àCl- + 6OH-   , Eo= 0.622V   (2)

Reversin the reaction gives Cl- + 6OH- -àClO3 (aq)+ 3H2O(l) Eo=-0.622 (2A)

Eq. 1+Eq.2A gives

3O3 +3H2O+Cl- + 6OH- --à3O2+6OH- +ClO3-(aq)+3H2O(l), Eo= 1.246-0.622 =0.624Vo

3O3+ Cl- -----à3O2+ClO3- Eo=0.624V

deltaGo=-nFE, n= no of electrons, F= 96500 coulumbs, E= 0.624V

deltaGo=-6*0.624*96500=-361296 Joules/mole= -361.3 KJ/mole

since deltaGo= -RTlnK

lnK= 361296/(8.314*298)= 146, K= 2.55*1063

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