Answer:
The value is [tex]K = 8*10^{61}[/tex]
Explanation:
From the question we are told that
The equation is [tex]2 Cr + 3Pb^{2+} \to 3Pb + 2Cr^{3+}[/tex]
The temperature is [tex] T = 25^oC = 298 K [room \ temperature ][/tex]
The emf at standard condition is [tex]E^o_{cell} = 0.61 \ V[/tex]
Generally at the cathode
[tex]3Pb^{2+}(aq) + 6 e- --> 3Pb(s)[/tex]
At the anode
[tex]2Cr^{3+} + 6e^- \to 2Cr[/tex]
Generally for an electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as
[tex]G = n* F * E^o_{cell} [/tex]
Here n is the no of electron with value n = 6
F is the Faraday's constant with value 96487 J/V
=> [tex]G = 6 * 96487 * 0.61[/tex]
=> [tex]G = 3.5 *10^{5} \ J[/tex]
This Gibbs free energy can also be represented mathematically as
[tex]G = RTlogK[/tex]
Here R is the cell constant with value 8.314J/K
K is the equilibrium constant
From above
=> [tex]K = antilog^{\frac{G}{ RT} }[/tex]
Generally antilog = 2.718
=>[tex]K = 2.718^{\frac{3.5 *10^5}{ 8.314* 298} }[/tex]
=> [tex]K = 8*10^{61}[/tex]
