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 0.0006 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

Explanation / Answer

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

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