Electrochemistry - 3rd Feb'16

Question

In the following electrochemical cell
!!Zn|Zn^(2+) || H^(+)|Pt(H_2)!!

!!E_"cell"^0 = E_"cell".!! This will be when

a) !![Zn^(2+)]= [H^+] = 1M!! and !!p_(H_2) = 1!!atm

b) !![Zn^(2+)]=0.01M, [H^+] = 0.1M!! and !!p_(H_2) = 1!! atm

c) !![Zn^(2+)]=1M, [H^+] = 0.1M!! and !!p_(H_2) = 1 !!atm

d) None of the above

Solution

Cell reaction:

!!Zn(s)+ 2H^+ (aq.) -> Zn^(2+)(aq.) +H_2 (g)!!

!!E_"cell" = E_"cell" ^0 - 0.0591/2 log_10 [(p_(H_2)[Zn^(2+)])/[H^+]^2]!!

!!Q_"cell" = (p_(H_2)[Zn^(2+)])/[H^+]^2!!

For !!E_"cell"^0 = E_"cell", !!Q_"cell" = 1!!

Option (A), !!Q_"cell" =(1xx1)/1^2=1!!

Option (B), !!Q_"cell" = (1xx0.01)/(0.1)^2=1!!

Option (C), !!Q_"cell" = (1xx1)/(0.1)^2 = 100!!

Therefore, option (A) and (B) are correct.