Predict cell potential under nonstandard conditions what is the calculated value

Question

Predict cell potential under nonstandard conditions what is the calculated value of the cell potential at 298K for an electrochemical cell with the reaction shown below when the Ag 1.20 M and the Mg -6398104 M? 2Ag (aq) + Mg(s) 2Ag(sM(aq) Cell Potential Use measured cell potential to calculate concentration. When [Hg21-1.12 M, the observed cell potential at 298K for an electrochemical cell with the reaction shown below is 2.127 V. What is the Mn2 concenation in this cell? Hg (aq) Mn(s) Hg(l) Mn(aq) [Mn2+] =

Explanation / Answer

1) 2 Ag+(aq) + Mg(s) ----> 2Ag(S) + Mg2+(aq)

Mg2+ + 2 e? <----> Mg(s)    E0 = ?2.372 v   

Ag+ + e? <-----> Ag(s)    E0 = +0.8 v

E0cell = E0cathode - E0anode

       = 0.8 - (-2.372)

       = 3.172 v

Ecell = E0cell - 0.0591/nlog(Mg2+/(Ag+)^2)

      = 3.172 - (0.0591/2)log((6.39*10^-4)/(1.2^2))

      = 3.27 v

2) Hg2+(aq) + Mn(s) -----> Hg(l) + Mn2+(aq)

   E0cell = E0cathode - E0anode

          = 0.85 - (-1.185)

          = 2.035 V

Ecell = E0cell - 0.0591/nlog([Mn2+]/[Hg2+])

   2.127 = 2.035 - (0.0591/2)log(x/1.12)

    x = [Mn2+]= 8.63*10^-4 M

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