Using data found in Appendix E of your textbook calculate the nonstandard emf fo

Question

Using data found in Appendix E of your textbook calculate the nonstandard emf for each of the following reactions if the concentration of each of the ions in these reactions is 0.0002 molar and everything else is standard (use 298 K for the temperature, R = 8.314 J/mol-K, and F = 96,485 C/mol):

(a) 1 Ca2+(aq) + 1 Sn2+(aq) --> 1 Ca(s) + 1 Sn4+(aq)

E = _____ V

(b) 1 Li+(aq) + 1 Fe2+(aq) --> 1 Li(s) + 1 Fe3+(aq)

E = _____ V

(c) 1 Hg22+(aq) + 2 Cl-(aq) --> 2 Hg(l) + 1 Cl2(g)

E = _____ V

(d) 2 Al3+(aq) + 3 Zn(s) --> 2 Al(s) + 3 Zn2+(aq)

E = _____ V

Explanation / Answer

From Nernst Equation we have

Ecell = Eocell -0.0592/n log Q

(a) 1 Ca2+(aq) + 1 Sn2+(aq) --> 1 Ca(s) + 1 Sn4+(aq)

Eocell =  E°cat - E°an   

ANODE: oxidation

CATHODE: reduction

Eocell =-2.76 + 0.15 = -2.61

Ecell = Eocell -0.0592/n log Q

Ecell = -2.61 - 0.0592/2 x log 0.0002 /0.0002 x 0.0002

Ecell = -2.61 - 0.109

Ecell = -2.719V

(b) 1 Li+(aq) + 1 Fe2+(aq) --> 1 Li(s) + 1 Fe3+(aq)

Eocell =  E°cat - E°an   

Eocell =-3.05 + 0.77 = -2.28

Ecell = Eocell -0.0592/n log Q

Ecell = -2.28 - 0.0592/1 x log 0.0002 / 0.0002 x 0.0002

Ecell = -2.28 - 0.219

Ecell = -2.499 V

3) 1 Hg22+(aq) + 2 Cl-(aq) --> 2 Hg(l) + 1 Cl2(g)

Eocell =  E°cat - E°an   

Eocell =0.27 + 1.36 = 1.63

Ecell = Eocell -0.0592/n log Q

Ecell = 1.63 - 0.0592/2 x log 0.0002 / 0.0002

Ecell = 1.63 - 0

Ecell =1.63 V

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