Question

Reduction half-reaction	E∘ (V)
Ag+(aq)+e−→Ag(s)	0.80
Cu2+(aq)+2e−→Cu(s)	0.34
Sn4+(aq)+4e−→Sn(s)	0.15
2H+(aq)+2e−→H2(g)	0
Ni2+(aq)+2e−→Ni(s)	−0.26
Fe2+(aq)+2e−→Fe(s)	−0.45
Zn2+(aq)+2e−→Zn(s)	−0.76
Al3+(aq)+3e−→Al(s)	−1.66
Mg2+(aq)+2e−→Mg(s)	−2.37

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 = 3.80×10⁻⁴.

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

Answer 1

1)   Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s)

     Fe(s) → Fe2+(aq) + 2e-          E^o = +0.45 V

      Ni2+ + 2e- ---------> Ni             E^o = -0.26 V

           E^ocell = +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 =  nFE^o / 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 = 3.80×10⁻⁴

    no of electrons involved n=1

From above,

   InK =  nFE^o / RT

Eo = [RTInK] / nF

        = 8.314 J/K/mol x 298 K x In (3.80×10⁻⁴)/ 1 x 96500 C/mol

        = + 0.087 V

   Eo = + 0.087 V