Question

ZnCl2 (aq) + Al (s) ---- Zn(s) + AlCl3 (aq)

[Al3+] = 0.8 M    [Zn2+] = 1.3M

Zn2+(aq) +2e- ---Zn(s)    -0.76V

Al3+(aq) + 3e- --- Al (s)    -1.66V

Balance the equation above.

Give the equation for the reaction quotient(Q) and solve for it. Are there units for (Q)?

Solve for the standard potential for the cell (EO cell) at 295K. First give equation.

Solve the Nernst equation for this cell at 295K.

Answer 1

The balanced equation is    : 3ZnCl₂ (aq) + 2Al (s) 3Zn(s) + 2AlCl₃ (aq)

Reaction quotient , Q = [AlCl₃ ] / [ZnCl₂]

                                  = 0.8 M / 1.3 M

                                 = 0.61

There are no units for Q.

Zn²⁺(aq) +2e^-Zn(s)    -0.76V     ( at cathode )     ----(1)

Al³⁺(aq) + 3e^-Al (s)    -1.66V    ( at anode )      ------(2)

The balanced equation is obtained as follows :

[3xEqn(1)] + reverse of [2xEqn(2)] gives

3Zn²⁺(aq) +6e^- +2Al 3Zn(s) +2Al³⁺(aq) + 6e^-

  3Zn²⁺(aq) +2Al 3Zn(s) +2Al³⁺(aq) ----(3)

So the number of electrons transferred are 6

The reduction potential of Al is less it acts as anodic reaction.Zn acts as cathodic reaction

So E^o = E _cathode - E _anode

          = E _Zn2+/Zn - E_{Al3+ / Al}

          = -0.76 - (-1.66) V

          = +0.90 V

According to Nernst Equation ,

E_o^cell = E^o - (0.059/n) log Q

         = +0.90 - (0.059/6) log 0.61

        = +0.89 V