1) Use the table of standard reduction potentials given above to calculate the e

Question

1)

Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 C) for the following reaction: Fe(s)+Ni2+(aq)Fe2+(aq)+Ni(s)

2)

Calculate the standard cell potential (E) for the reaction

X(s)+Y+(aq)X+(aq)+Y(s)

if K = 3.80×104.

Express your answer to three significant figures and include the appropriate units.

Reduction half-reaction E (V) Ag+(aq)+eAg(s) 0.80 Cu2+(aq)+2eCu(s) 0.34 Sn4+(aq)+4eSn(s) 0.15 2H+(aq)+2eH2(g) 0 Ni2+(aq)+2eNi(s) 0.26 Fe2+(aq)+2eFe(s) 0.45 Zn2+(aq)+2eZn(s) 0.76 Al3+(aq)+3eAl(s) 1.66 Mg2+(aq)+2eMg(s) 2.37

Explanation / Answer

1)   Fe(s)+Ni2+(aq)Fe2+(aq)+Ni(s)

     Fe(s) Fe2+(aq) + 2e-          Eo = +0.45 V

      Ni2+ + 2e- ---------> Ni             Eo = -0.26 V

           Eocell = +0.45 V-0.26 V = + 0.19 V

         no of electrons involved , n =2

       We know that    Go = -nFEo and    Go = -RT InK

           Hence,

           -nFEo = -RT InK

             InK =  nFEo / RT

             K = e^(nFEo/RT)

                = e^(2x 96500 C/mol x 0.19 V/ 8.314 J/K/mol x 298 K)

                = 2678535

            K = 2678535

Therefore, equilibrium constant = 2678535

2)  

X(s)+Y+(aq)X+(aq)+Y(s)                  K = 3.80×104

    no of electrons involved n=1

From above,

   InK =  nFEo / RT

Eo = [RTInK] / nF

        = 8.314 J/K/mol x 298 K x In (3.80×104)/ 1 x 96500 C/mol

        = + 0.087 V

   Eo = + 0.087 V

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