Electrochemistry

Consider the following 4 electrodes

$$\begin{aligned} & \mathrm{A}: \mathrm{Ag}^{+}(0.0001 \mathrm{M}) / \mathrm{Ag}(s) ; \\ & \mathrm{B}: \mathrm{Ag}^{+}(0.1 \mathrm{M}) / \mathrm{Ag}(s) ; \\ & \mathrm{C}: \mathrm{Ag}^{+}(0.01 \mathrm{M}) / \mathrm{Ag}(s) ; \\ & \mathrm{D}: \mathrm{Ag}^{+}(0.001 \mathrm{M}) / \mathrm{Ag}(s) ; E^{\circ}{ }_{\mathrm{Ag}^{+} / \mathrm{Ag}}=+0.80 \mathrm{~V} \end{aligned}$$

Then reduction potential in volts of the electrodes in the order.