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 0.0006 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 Cl2(g) + 1 Ca(s) --> 2 Cl-(aq) + 1 Ca2+(aq) E = ___V

b) 2 Cu2+(aq) + 1 Fe(s) --> 2 Cu+(aq) + 1 Fe2+(aq) E = ___V

(c) 1 Hg22+(aq) + 2 Cu(s) --> 2 Hg(l) + 2 Cu+(aq) E =___ V

(d) 1 Cd2+(aq) + 1 Sn(s) --> 1 Cd(s) + 1 Sn2+(aq) E = ___V

Answer 1

We will use Nernst equation to calculate the Ecell

a) we know that

Ecell = E⁰cell - 0.0592 / n log Q

E⁰cell = E⁰cathode - E⁰anode

E⁰cell = 1.36 - (-2.76)

E⁰cell = 4.12

log Q = log [Ca+2] / [Cl-]²

log Q = log (1 / 0.0006)

Ecell = E⁰cell - 0.0592 /2 log (q)

Ecell = 4.12 - 0.095 = 4.025

b) we know that

Ecell = E⁰cell - 0.0592 / n log Q

E⁰cell = E⁰cathode - E⁰anode

E⁰cell = 0.16 - (-0.41))

E⁰cell = 0.57

log Q = log [Fe+2] [Cu⁺] ²/ [Cu⁺²]²

log Q = log 0.0006

Ecell = E⁰cell - 0.0592 /2 log (q)

Ecell =0.57 + 0.095 = 0.665

c) we know that

Ecell = E⁰cell - 0.0592 / n log Q

E⁰cell = E⁰cathode - E⁰anode

E⁰cell = 0.8 - 0.52

E⁰cell = 0.28

log Q = log [Cu⁺] ²/ [Hg2⁺²]

log Q = log 0.0006

Ecell = E⁰cell - 0.0592 /2 log (q)

Ecell =0.28 + 0.095 = 0.375

d) we know that

Ecell = E⁰cell - 0.0592 / n log Q

E⁰cell = E⁰cathode - E⁰anode

E⁰cell = -0.4 - (-0.14)

E⁰cell = -0.26

log Q = log [Sn+2]/ [Cd⁺²]

log Q = log 1

Ecell = E⁰cell - 0.0592 /2 log (q)

Ecell = -0.26 V