You have a concentration cell. The half-cell reactions are the same, but the concentrations of the ions are different.
The half-cell reactions are
color(white)(mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmll)ul(E^@"/V")
"Anode:"color(white)(mll) "Cu"color(white)(mmmmmmmmmm) ⇌ "Cu"^"2+"("0.001 mol/L") + color(red)(cancel(color(black)("2e"^"-")))color(white)(m)"+0.334"
"Cathode:"color(white)(l) ul("Cu"^"2+"("0.1 mol/L") + color(red)(cancel(color(black)("2e"^"-"))) ⇌ "Cu"color(white)(mmmmmmmmmmm))color(white)(mll)ul("-0.334")
"Overall:"color(white)(m) "Cu" + "Cu"^"2+"("0.1 mol/L") ⇌ "Cu"^"2+"("0.001 mol/L") + "Cu"color(white)(mmll)0
The Nernst equation for the cell potential is
color(blue)(bar(ul(|color(white)(a/a)E = E_text(cell)^° - (RT)/(zF)lnQcolor(white)(a/a)|)))" "
where
E_text(cell)^@ = the standard cell potential
Rcolor(white)(ml) = the Universal Gas Constant
T color(white)(ml)= the temperature
zcolor(white)(mll) = the moles of electrons transferred per mole of Cu
F color(white)(ml) = the Faraday constant
Q color(white)(ml) = the reaction quotient
Calculate the cell potential
Assume that the temperature is 25 °C.
E_text(cell)^@ = 0
Q = (["Cu"^"2+"]_text(prod))/(["Cu"^"2+"]_text(react)) = (0.001 color(red)(cancel(color(black)("mol/L"))))/(0.1 color(red)(cancel(color(black)("mol/L")))) = 0.01
E = E_text(cell)^@ -(RT)/(zF)lnQ = 0 -("8.314 V"·color(red)(cancel(color(black)("C·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K"))))/("2 × 96 485" color(red)(cancel(color(black)("C·mol"^"-1"))))ln(0.01) = "0.059 V"
