Question

Question 7 of 16 > Calculate the cell potential, E, for the given reactions at 25.00 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (E ). Pt(s) + Fe2+ (aq) = P22+ (aq) + Fe(s) [Fe2+] = 0.0017 M [P12+] = 0.040 M The cell is O spontaneous. O not spontaneous

Answer 1

Pt(s) ------------> Pt^2+ (aq) +2e^-           E0   = -1.20v

Fe^2+ (aq) + 2e^- ------> Fe(s)                E0   = -0.41v

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Pt(s) + Fe^2+ (aq) -----> Pt^2+ (aq) + Fe(s)    E0cell = -1.61v

E0cell   = -1.61v

n   = 2

E   = E0cell - 0.0592/n logQ

     = -1.61-0.0592/2log[Pt^2+]/[Fe^2+]

      = -1.61-0.0296log(0.04/0.0017)

      =-1.61-0.029*1.3716

      = -1.65v

E0cell<0

It is non spontaneous

Cu(s) --------> Cu^2+ (aq) + 2e^-        E0   = -0.34v

2Ag^+ (aq) + 2e^- ------> 2Ag(s)         E0   = 0.80v

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Cu(s) + 2Ag^+ (aq) -----> Cu^2+ (aq) + 2Ag(s)    E0cell = 0.46v

n = 2

E   = E0cell - 0.0592/n logQ

     = 0.46-0.0592/2log[Cu^2+]/[Ag^+]^2

     = 0.46-0.0296log(0.021/(0.021)^2

      = 0.46 -0.0296*1.678

      = 0.41v

E0cell>0

It is spontaneous

Co^2+ (aq) ----------------> Co^3+ (aq) + e^-            E0 = -1.82v

Ti^3+ (aq) + e^- -----------------> Ti^2+ (aq)             E0   = -0.90v

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Co^2+ (aq) + Ti^3+ (aq) ------> Co^3+ (aq) + Ti^2+ (aq)    E0cell = -2.72v

n = 1

E   = E0cell - 0.0592/n logQ

      = -2.72 - 0.0592/1log[Co^3+][Ti^2+]/[Co^2+][Ti^3+]

     = -2.72-0.0592 log(0.04*0.0098/0.05*0.0045)

     =-2.72 -0.0592*0.241

       = -2.734v

E0cell<0

it is non spontaneous