This is a mis - leading question as the -ve E_(cell)^@ value tells us that the reaction cannot happen as written left to right under standard conditions.
For a reaction to occur the free energy change must be -ve:
DeltaG=-nFE_(cell)
This means E_(cell) must be positive for this to occur. In this case we are told E_(cell)^@=-0.271"V".
However, we are not under standard conditions since pH=5.6 and under standard conditions pH=0.
So we need to use The Nernst Equation:
E_(cell)=E_(cell)^@-0.05916/(n)logQ" "color(red)((1))
(At 25^@"C")
NH_(4(aq))^(+)+2O_(2(g))+H_2O_((l))rarrNO_(3(aq))^(-)+2H_((aq))^(+)
We are told the reaction has reached equilibrium. At this point the potential difference between the 2 half - cells has fallen to zero.
So color(red)((1)) becomes:
0=E_(cell)^(@)-0.05916/(n)logK
Note we have replaced Q with K.
:.0=-0.271-0.05916/(8)logK
n=8 which is the number of moles of electrons transferred.
:.-0.05916/(8)logK=0.271
:.logK=-(0.271xx8)/(0.05916)
logK=-36.64
:.K=2.29xx10^(-37)
K=([NO_3^-][H^(+)]^(2))/([NH_4^(+)]pO_2^(2))" "color(red)((2))
We can find [H^(+)] since pH=5.6
:.-log[H^(+)]=5.6
:.[H^+]=2.512xx10^(-6)"mol/l"
We are told pO_2=0.18"Atm". We can use this in the expression since the values used for K are normalised against standard conditions for which p=1"Atm" and K is dimensionless.
Now we can put values into color(red)((2))rArr
2.29xx10^(-37)=([NO_3^-])/([NH_4^+])xx((2.512xx10^(-6))^2)/(0.18^2)
:.([NO_3^-])/([NH_4^+])=1.175xx10^(-27)
This, and the tiny value of K, tells us that the equilibrium lies almost completely to the left and the equation should more correctly written right to left.
Another mis - leading aspect of what I consider to be a badly written question is that the E^@=1.5"V" value given is not needed.
