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#