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**Hint:**Try to find the concentration of \[Z{{n}^{2+}}\] which is to be used in the reaction Quotient. Electrode potential can be calculated using the Nernst equation.

**Complete answer:**

Electrode potential refers to the potential difference developed across the electrical double layer in one half cell. In simple words, it refers to the emf of half cell reaction.

The absolute value of the electrode potential cannot be measured as there is no flow of electrons through wire across the electrical double layer.

Let’s find the electrode potential in this question,

Dissociation of \[ZnS{{O}_{4}}\]:

\[ZnS{{O}_{4}}\left( s \right)\to Z{{n}^{2+}}\left( aq \right)+S{{O}_{4}}^{2-}\left( aq \right)\]

As it is given that the \[0.1M\]solution of \[ZnS{{O}_{4}}\] does not dissociate fully, It is dissociating only \[95\%\], therefore the concentration of \[Z{{n}^{2+}}\left( aq \right)\]will be \[0.1M\times \dfrac{95}{100}=0.095M\]

Now let’s write the half cell reaction of which we are supposed to find the electrode potential

\[Z{{n}^{2+}}\left( aq \right)+2{{e}^{-}}\to Zn\left( s \right)\]

The equation to be used is Nernst equation, which is

\[E=\left[ {{E}^{\circ }}-\left( \dfrac{0.0591}{n} \right){{\log }_{10}}Q \right]\]

\[{{E}^{\circ }}=\] Standard emf of the reaction

\[n=\] The n-value for the cell reaction is the number of moles of electrons released at anode or the number of moles of electrons gained at the cathode.

\[Q=\] Reaction quotient. It has the same expression as that of the equilibrium constant but it can be used anytime.

Now, the n-value for the half cell reaction is \[2\]in our case as two electrons are gained at the cathode.

Reaction quotient for the half cell reaction is

\[\dfrac{1}{\left[ Z{{n}^{2+}}(aq) \right]}=\dfrac{1}{0.095}\]

Pure solid \[Zn\left( s \right)\]does not appear in the reaction quotient.

\[{{E}^{\circ }}\]is already given to us which is \[-0.76V\].

Now, putting all these values in the nernst equation, we get

\[\begin{align}

& \rightarrow E=-0.76-\dfrac{0.0591}{2}\times \log \dfrac{1}{0.095} \\

& \rightarrow E=-0.76-\dfrac{0.0591}{2}\times (\log 1000-\log 95) \\

& \rightarrow E=-0.76-\dfrac{0.0591}{2}\times (3-1.9777) \\

& \rightarrow E=-0.76-0.0302 \\

& E=-0.7902V \\

\end{align}\]

**Hence, Electrode potential is\[-0.7902V\]**

**Note:**

Electrode potential is of two types:

-Oxidation potential

-Reduction Potential

The greater the oxidation potential, the more is the tendency to be oxidised. A stronger reducing agent has high oxidation potential.

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