Determine equilibrium constants for the following reactions. (See this appendix.

Question

Determine equilibrium constants for the following reactions. (See this appendix.)

a. The half-cell potentials for the O2/H2O and the H2O/H2 systems are for neutral (pH = 7) water and are not standard reduction potentials where [OH1-] or [H1+] = 1.0 M.

Reduction half-reactions Eo(V) Reduction half-reactions Eo(V) Li1+(aq) + e1- Li(s) -3.04 Cu2+(aq) + 2e1- Cu(s) +0.34 K1+ (aq)+ e1- K(s) -2.92 Ag2O(s) +H2O + 2e1- 2Ag(s) + 2OH1-(aq) +0.34 Ba2+(aq) + 2e1- Ba(s) -2.92 ClO31-(aq) + H2O(l) + 2e1- ClO21-(aq) + 2OH1-(aq) +0.35 Ca2+(aq) + e1- Ca(s) -2.84 O2(g) + 2H2O + 4e1- 4OH1-(aq) +0.40 Na1+(aq) + e1- Na(s) -2.71 I2(s) + 2e1- 2I1-(aq) +0.54 Mg2+(aq) + 2e1- Mg(s) -2.36 ClO21-(aq) + H2O(l) + 2e1- ClO1-(aq) + 2OH1-(aq) +0.59 Al3+(aq) + 3e1- Al(s) -1.66 2AgO(s) + H2O + 2e1- Ag2O(s) + 2OH1-(aq) +0.60 U3+(aq) + 3e1- U(s) -1.66 O2(g) + 2H1+(aq) + 2e1- H2O2(aq) +0.70 Ti2+(aq) + 2e1- Ti(s) -1.63 Fe3+(aq) + e1- Fe2+(aq) +0.77 Mn2+(aq) + 2e1- Mn(s) -1.18 BrO1-(aq) + H2O + 2e1- Br1-(aq) + 2OH1-(aq) +0.77 2H2O + 2e1- H2(g) + 2OH1-(aq) -0.83 Ag1+(aq) + e1- Ag(s) +0.80 Zn2+(aq) + 2e1- Zn(s) -0.76 O2(g)+ 4H1+(aq) + 4e1- 2H2O +0.82* Cr3+(aq) + 3e1- Cr(s) -0.74 H2O2(aq) + 2e1- 2OH1-(aq) +0.88 HCHO(aq) + 2H2O + 2e1- CH3OH(aq) + 2OH1-(aq) -0.59 ClO1-(aq) + H2O + 2e1- Cl1-(aq) + 2OH1-(aq) +0.89 Fe2+(aq) + 2e1- Fe(s) -0.44 Hg2+(aq) + 2e1- Hg(l) +0.85 2H2O + 2e1- H2(g) + 2OH1-(aq) -0.41 * NO31-(aq) + 4H1+(aq) + 3e1- NO(g) + 2H2O +0.96 Cd2+(aq) + 2e1- Cd(s) -0.40 VO21+(aq) + 2H1+(aq) + e1- VO2+(aq) + H2O +1.00 PbSO4(s) + 2e1- Pb(s) + SO42-(aq) -0.36 Br2(l) + 2e1- 2Br1-(aq) +1.09 In3+(aq) + 3e1- In(s) -0.34 ClO41-(aq) + 2H1+(aq) + 2e1- ClO31-(aq) + H2O(l) +1.19 Co2+(aq) + 2e1- Co(s) -0.28 O2(g)+ 4H1+(aq) + 4e1- 2H2O +1.23 Ni2+(aq) + 2e1- Ni(s) -0.23 Cr2O72-(aq)+ 14H1+(aq) + 6e1- 2Cr3+(aq) + 7H2O +1.33 Sn2+(aq) + 2e1- Sn(s) -0.14 Cl2(g) + 2e1- 2Cl1-(aq) +1.36 Pb2+(aq) + 2e1- Pb(s) -0.13 Au3+(aq) + 3e1- Au(s) +1.50 2H1+(aq) + 2e1- H2(g) 0.00 MnO41-(aq) + 8H1+(aq)+ 5e1- Mn2+(aq) + 4H2O +1.51 Sn4+(aq) + 2e1- Sn2+(aq) +0.15 PbO2(s) + 4H1+(aq) + SO42-(aq) + 2e1- PbSO4(s) + 2H2O +1.69 Cu2+(aq) + e1-` Cu1+(aq) +0.16 H2O2(aq) + 2H1+(aq) + 2e1- 2H2O(l) +1.76 ClO41-(aq) + H2O(l) + 2e1- ClO31-(aq) + 2OH1-(aq) +0.17 S2O82-(aq) + 2e1- 2SO42-(aq) +2.01 AgCl(s) + e1- Ag(s) + Cl1-(aq) +0.22 O3(g) + 2H1+(aq) + 2e1- O2(g) + H2O +2.07 PbO2(s) + 2H1+(aq) + 2e1- PbO(s) + H2O +0.28 F2(g) + 2e1- 2F1-(aq) +2.87

Explanation / Answer

from the given data

E0cell = Ecathode - Eanode

        = 2.07 - 1.36

        = 0.71 V

DG0 = -nFE0cell

     = -2*96500*0.71 = -137030 joule.

dG0 = -RTlnK

    -137030 = -8.314*273lnk

k = 1.66*10^26

b)

from the given data

E0cell = Ecathode - Eanode

        = -0.13 - -0.14

        = 0.01 V

DG0 = -nFE0cell

     = -2*96500*0.01 = -1930 joule.

dG0 = -RTlnK

    -1930 = -8.314*273lnk

k = 2.34

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