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**Hint:**In an electrolytic cell when a metal electrode is placed in solution of its ions of unit molarity then the potential difference generated between the electrode and the ions in solutions is called as standard electrode potential.

**Complete step by step answer:**

According to the IUPAC convention, standard reduction potential is taken as the electrode potential. The given value of electrode potential can be considered as the standard reduction potential unless it is mentioned that it is oxidation potential.

Standard reduction potential means that reduction is taking place at the electrode. Also standard oxidation has the same value as the standard reduction potential only the sign is reversed. Thus,

Standard oxidation potential$ = - $Standard reduction potential

Consider the given example,

The electrode for $C{u^{2 + }}/Cu$ couple can be given as-

$C{u^{2 + }} + 2{e^ - } \to C{u_{(s)}}$

The electrode potential for the electrode is given $E_{red}^\circ = 0.34V$ . Standard electrode potential is potential when the solution is of unit molarity.

Now when the concentration of $C{u^{2 + }}$ is $0.01M$ the electrode is not equal to standard electrode potential since concentration changes. We can measure the electrode potential with respect to standard hydrogen electrodes(SHE). SHE is used as a reference electrode because its potential is considered to be zero. So the cell potential will be equal to the electrode potential. So the electrode potential can be calculated by using Nernst equation-

$E_{C{u^{2 + }}/Cu}^{} = E_{C{u^{2 + }}/Cu}^\circ - \dfrac{{2.303RT}}{{nF}}\log \dfrac{{[product]}}{{[reactan t]}}$

$n$ is the number of electrons required for reduction, $R$ is gas constant, $T$ is the temperature, $F$ is Faraday’s constant, $[product]$ is the concentration of product, $[reac\tan t]$ is the concentration of reactant, $E_{C{u^{2 + }}/Cu}^\circ $ is the standard reduction potential of copper electrode and ${E_{C{u^{2 + }}/Cu}}$ is the electrode potential at the given concentration.

At $298K$ , $\dfrac{{2.303RT}}{F} = 0.059$

So, $E_{C{u^{2 + }}/Cu}^{} = E_{C{u^{2 + }}/Cu}^\circ - \dfrac{{0.059}}{n}\log \dfrac{{[product]}}{{[reactan t]}}$

Now for the half-cell reaction,

$C{u^{2 + }} + 2{e^ - } \to C{u_{(s)}}$

$E_{C{u^{2 + }}/Cu}^{} = E_{C{u^{2 + }}/Cu}^\circ - \dfrac{{0.059}}{n}\log \dfrac{{[product]}}{{[reactan t]}}$

$n = 2$, \[E_{C{u^{2 + }}/Cu}^\circ = 0.34V\] , $[C{u^{2 + }}] = 0.01M$ , $[Cu] = 1$ since activity of metal is taken as unity.

$E_{C{u^{2 + }}/Cu}^{} = E_{C{u^{2 + }}/Cu}^\circ - \dfrac{{0.059}}{n}\log \dfrac{{[Cu]}}{{[C{u^{2 + }}]}}$

\[E_{C{u^{2 + }}/Cu}^{} = 0.34 - \dfrac{{0.059}}{2}\log \dfrac{{[1]}}{{[0.01]}}\]

${E_{C{u^{2 + }}/Cu}} = 0.34 - 0.0295\log ({10^2})$

${E_{C{u^{2 + }}/Cu}} = 0.281V$

**So, the correct answer is Option B.**

**Note:**HE is used as a primary reference electrode to find out the electrode potential of other elements. When ${E^\circ }$ is positive the reaction is spontaneous. $E_{cell}^\circ = $ standard reduction potential of cathode $ - $ standard reduction potential of anode

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