Consider the following four electrodes: 
$\begin{array}{l}\mathrm{A}={\mathrm{Cu}}^{2+}\left(0.0001\mathrm{M}\right)/\mathrm{Cu}\left(\mathrm{s}\right)\\ \mathrm{B}={\mathrm{Cu}}^{2+}\left(0.1\mathrm{M}\right)/\mathrm{Cu}\\ \mathrm{C}={\mathrm{Cu}}^{2+}\left(0.01\mathrm{M}\right)/Cu\\ \\ \mathrm{D}={\mathrm{Cu}}^{2+}\left(0.001\mathrm{M}\right)/{\mathrm{Cu}}_{\left(\mathrm{s}\right)}\end{array}$
If the standard reduction potential of Cu is + 0.34V, the reduction potentials (in volts) of the above electrodes follow the order

The Nernst equation is written as 

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}+\hspace{0.17em}\frac{0.0591}{\mathrm{n}}\log \left[{\mathrm{Cu}}^{2+}\right]$

Lower the concentration of [Cu²⁺] , Lower is the value of E_cell. So, B has the highest reduction potential value because of high concentration of Cu²⁺

The correct trend is B>C>D>A