Using data found in Appendix E of your textbook calculate the nonstandard emf fo
ID: 513745 • Letter: U
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 0.0008 molar and everything else is standard (use 298 K for the temperature R = 8.314 J/mol-K, and F = 96, 485 C/mo (a) 2 Fe(CN_6)^3-(aq) + 1 Sn^2+(aq) rightarrow 2 Fe(CN_6)^4-(aq) + 1 Sn^4+(aq) E = V (b) 1 Mg^2+(aq) + 1 Ni(s) rightarrow 1 Mg(s) + 1 Ni^2+(aq) E = V (c) 3 Na^+(aq) + 1 Al(s) rightarrow 3 Na(s) + 1 Al^3+(aq) E = V (d) 1 Fe^3+(aq) + 1 Cr^2+(aq) rightarrow 1 Fe^2(aq) + 1 Cr^3+(aq) E = VExplanation / Answer
a)
Fe goes from +3 to +2
Fe3+ + e Fe2+ +0.77
Sn2+ Sn4+ + 2 e -0.15
E° = 0.77+-0.15 = 0.62
E = E° - 0.0592/n*log(Q)
n = 2 electrons, Q = [Sn+4][Fe+2]/[Fe+3][Sn+2]
E = 0.62 - 0.0592/(2)*log(0.0008*0.0008^2/(0.0008^2)(0.0008))
E = 0.62
B)
Mg2+ + 2 e Mg(s) 2.372
Ni2+ + 2 e Ni(s) 0.25
E° = -0.25 + 2.372 = 2.122
E = E° - 0.5092/n * log(Q)
Q = [Mg+2]/[Ni+2] = 0.0008/0.008
E = 2.122 V
c)
Na+ + e Na(s) 2.71
Al3+ + 3 e Al(s) 1.662
E° = -1.662 - -2.71 = 1.048 V
E = E° -0.0592/n * log(Q)
E = 1.048-0.0592/3*log(0.0008/(0.0008^3)
E = 0.925 V
D)
Fe3+ + e Fe2+ +0.77
Cr2+ Cr3+ + e 0.42
E° = 0.77+0.42 = 1.19
E = E° -0.0592/n * log[Fe+2][Cr+3]/([Cr+2][Fe+3])
E = 1.19-0.0592/1 * log(0.008*0.008/0.008/0.008)
E =1.19