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"#
