Question

Calculate the equilibrium constant for the reaction: 2 Cr + 3 Pb2+ ----> 3 Pb + 2 Cr3+ at 25oC. Eocell = 0.61 V

Answer 1

Answer:

The value is  $K = 8*10^{61}$

Explanation:

From the question we are told that

    The equation is  $2 Cr + 3Pb^{2+} \to 3Pb + 2Cr^{3+}$

     The  temperature is  $T = 25^oC = 298 K [room \ temperature ]$

     The  emf at standard condition is  $E^o_{cell} = 0.61 \ V$

Generally at the cathode

      $3Pb^{2+}(aq) + 6 e- --> 3Pb(s)$

  At the anode

      $2Cr^{3+} + 6e^- \to 2Cr$

Generally for an  electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as  

       $G = n* F * E^o_{cell}$

Here  n  is  the no of electron  with value n = 6

       F  is  the Faraday's constant with value 96487 J/V

  =>   $G = 6 * 96487 * 0.61$

  =>   $G = 3.5 *10^{5} \ J$

This Gibbs free energy can also be represented mathematically as

       $G = RTlogK$

Here  R  is the cell constant with value 8.314J/K

           K is the equilibrium constant

From above

=>  $K = antilog^{\frac{G}{ RT} }$

Generally  antilog =  2.718

=>$K = 2.718^{\frac{3.5 *10^5}{ 8.314* 298} }$

=>   $K = 8*10^{61}$