$\begin{array}{1 1}(a)\;-0.79Volt\\(b)\;0.79Volt\\(c)\;-0.89Volt\\(d)\;0.89Volt\end{array}$

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Concentration of $ZnSO_4$ solution =0.1M

$\%$ of dissociation of $ZnSO_4$ solution =95%

Concentration of $Zn^{+2}$ ions in $ZnSO_4$ solution =$\large\frac{0.1\times 95}{100}$

$\Rightarrow 0.095M$

Thus,the electrode can be represented as $Zn(s) /Zn^{+2}(0.095M)$

Reduction reaction taking place at this electrode is

$Zn^{+2}+2e^-\rightarrow Zn(s)$(Here n=2)

According to Nernst equation,the reduction potential of above electrode is given by

$E_{Zn^{+2}/Zn}=E^0_{Zn^{+2}/Zn}-\large\frac{0.059}{n}$$\log \large\frac{1}{0.095}$

$E_{Zn^{+2}/Zn}=-0.76+\large\frac{0.059}{2}$$\log 0.095$

$E_{Zn^{+2}/Zn}$=-0.79Volt

Hence (a) is the correct answer.

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