Question

1. The half reaction for a cell is given below.
                     Cu²⁺ + 2e à Cu_(s)

1. Calculate the equilibrium constant for the net reaction.
2. If there were a junction potential of +2mV (increasing E from 0.339 to 0.341 V), by what percentage would the calculate equilibrium constant increase?

Answer 1

a) The half reaction is given as

Cu²⁺ (aq) + 2 e^- ----------> Cu (s)

E⁰ = +0.339 V

The standard reduction potential is related to the equilibrium constant as

nFE⁰ = RTln K

where n is the number of electron(s) transferred = 2; F = 96485 J/V.mol is the Faraday’s constant, R = 8.314 J/mol.K is the gas constant and T = absolute temperature of the reaction = 298 K.

Therefore,

ln K = nFE⁰/RT

= (2)*(96485 J/V.mol)*(0.339 V)/(8.314 J/mol.K)(298 K)

= 26.4036

======> K = exp(26.4036)

======> K = 2.93*10¹¹

The equilibrium constant for the reaction under standard conditions is given as 2.93*10¹¹.

b) The standard reduction potential increases to 0.341 V.

Plug in values and get the value of the new equilibrium constant

ln K’ = nFE⁰’/RT

= (2)*(96485 J/V.mol)*(0.341 V)/(8.314 J/mol.K)(298 K)

= 26.5593

======> K’ = exp(26.5593)

======> K’ = 3.42*10¹¹

Percent increase in the value of the equilibrium constant

= (K’ – K)/K*100

= (3.42*10¹¹ – 2.93*10¹¹)/(2.93*10¹¹)*100

= 16.7% (ans).