The cell diagram
#color(white)(-----)color(red)["Zn"_((s)) | "Zn"^(2+) || "Ag"^(1+) | "Ag"_((s))#
tells us that the anode is on the left side of ||, and the cathode is to the right of ||.
Knowing this, we can figure out that #color(red)["Zn"_((s))# is going to be the one that is oxidized while #color(red)["Ag"^(+)# is going to be the one that is reduced. So the redox reaction would look something like this:
#color(white)(-----)color(red)["2Ag"^(+) + Zn_(s)rarr"2Ag"_((s)) + "Zn"^(2+)#
Seeing that we were given the standard reduction potentials, where the potentials are measured at #"298 K, 1 atm, and solutions at 1 M",# we can now go ahead and calculate the standard cell potential using the following equation:
#color(white)(------)color(blue)[E_"cell"^@ = E_"red"^@ + E_"ox"^@#
#color(white)(-----)#
#ul"Standard Reduction Potentials"#
#"Zn"^(2+) + 2e^"-" rarr "Zn"_((s))color(white)(---)E_"red"^@ = -0.76 V#
#"Ag"^(+) + 1e^"-" rarr "Ag"_((s))color(white)(---)E_"red"^@ = +0.80 V#
Note: Since we were given standard reduction potentials, in order to get the standard oxidation potential (#E_"ox"^@)# for zinc, we would just reverse the sign for the #E_"red"^@#. Consequently, we would show the reaction for zinc as being oxidized.
So,
#color(white)(--)E_"red"^@ = -0.76 Vcolor(white)(--)#becomes#color(white)(--)E_"ox"^@ = +0.76 V#
Solve for standard cell potential
#color(white)(--)color(blue)[E_"cell"^@ = E_"red"^@+E_"ox"^@]->(+0.80)+(+0.76)=barul (|+1.56 V |)#
#---------------------#
Now, we use something called the #"Nernst equation"# to calculate the non-standard cell potential, which I believe is what your question is asking.
#color(white)(-----)color(blue)[E_"cell" = E_"cell"^@-(0.059)/(n)logQ]#
Where
#"n = number of moles of electrons transferred (2)"#
#"Q = reaction quotient" [["Zn"^(2+)]]/[["Ag"^+]^2]#
#color(white)(--)#
Now, we would plugin where appropriate and solve(ignored units)
#color(blue)[E_"cell" = E_"cell"^@-(0.059)/(n)logQ]#
#E_"cell" =(+1.56)-((0.059))/((2))log([[0.001]]/[0.1]^2)#
#E_"cell" =(+1.56)-(-0.0295)#
#E_"cell" =color(magenta)[+1.58 V#
#"Answer": color(magenta)[+1.58 V#
