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.