Question

Consider the corrosion of copper at 25 C using the overall reaction Cu(s) + 0.5 O2(g) + 2 H3O+(aq) → Cu2+(aq) + 3 H2O(l) If the partial pressure of oxygen is 0.06 bar, and the concentration of Cu2+(aq) is 3 M, at what pH does the reaction become spontaneous?

Answer 1

Cu(s) + 0.5 O2(g) + 2 H3O+(aq) → Cu2+(aq) + 3 H2O(l)

P(O2)=0.06 bar

1bar=0.99 atm (approx)

Kp=1/pO2^0.5=1/(0.06)^0.5=4.08 atm^-1/2

Also Kp=Kc(RT)^∆n

R=gas constant=0.0821L atm/K mol

T=25+273=298K

∆n=no of moles of gaseous product-no of moles of gaseous reactants=0-0.5=-0.5

Or, kc=kp/(RT)^∆n=4.08 atm^-1/2/(0.0821L atm/K mol *298K)^(-0.5)=0.0824 mol/L or M

Kc=0.824 M=[Cu2+]/ [H3O+]^2 [activity of pure solid and liquid=1]

Given [Cu2+]=3M

0.0824 M=(3M)/ [H3O+]^2

Or, [H3O+]=0.524 M

pH=-log[H3O+]=-log 0.524=0.28

Equilibrium at pH=0.28 so spontaneous rxn, when [H3O+] conc is beyond this limit.(le chatlier’s principle)

or, pH<0.28