Reduction half-reaction Part A Use the table of standard reduction potentials gi

Question

Reduction half-reaction

Part A

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)

B) Calculate the standard cell potential (E?) for the reaction if K = 3.07

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

Explanation / Answer

Cell reactions

Fe(s) <---------------------> Fe2+(aq)+ 2e- , at anode E0= 0.45V

Ni2+(aq) + 2e-<---------------> Ni(s)            at cathode E0 = -0.26V

------------------------------------------------------------------------------------

Cell reaction Fe(s) + Ni2+(aq) <-----------------> Fe2+(aq) + Ni(s)      E0cell= 0.19V

We know RT lnK=nFE0cell

K= Equilibrium constant , n= number electrons transfered during the raection =2

T= 25C=298K, R= 8.314 J/mol. K , F= 96500 columes , E0cell= 0.19V

lnK=nFE0cell /RT

   = 2*96500 * 0.19 / 8.314 *298

   lnK = 14.8

K= 2.676*10^6

B). E0cell= (RT/nF) lnK

               = (8.314 *298/ 1* 96500) ln(3.07*10^-3)

               =0.0591 *(-2.513)

         E0cell      =- 0.1485V

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