Question

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

Answer 1

The value is
`K  =  8*10^(61)`

Step-by-step 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^{(G)/( RT) }`

Generally antilog = 2.718

=>
`K  =  2.718^{(3.5 *10^5)/( 8.314* 298) }`

=>
`K  =  8*10^(61)`