$$ \text { Consider the following data. } $$

$$ \begin{array}{|c|c|} \hline \text { Electrolyte } & \wedge^{\circ}_\mathbf{m}{\mathbf{(}} \mathbf{S ~ c m}^{\mathbf{2}} \mathbf{~ m o l}^{\mathbf{1}} \mathbf{)} \\ \hline \mathrm{BaCl}_2 & x_1 \\ \hline \mathrm{H}_2 \mathrm{SO}_4 & x_2 \\ \hline \mathrm{HCl} & x_3 \\ \hline \end{array} $$

$\mathrm{BaSO}_4$ is sparingly soluble in water. If the conductivity of the saturated $\mathrm{BaSO}_4$ solution is $x \mathrm{~S} \mathrm{~cm}^{-1}$ then the solubility product of $\mathrm{BaSO}_4$ can be given as (Here $\wedge_{\mathrm{m}}=\wedge^{\circ}_{\mathrm{m}}$ )

One half cell in a voltaic cell is constructed by dipping silver rod in $\mathrm{AgNO}_3$ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of $\mathrm{ZnSO}_4$.

A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of $\mathrm{Ag}^{+}$ions used in terms of $\log x\left(x=\left[\mathrm{Ag}^{+}\right]\right)$?

$$ \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\ominus}=-0.76 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\ominus}=+0.80 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.059 \mathrm{~V} $$

An electrochemical cell is constructed using half cells in the direction of spontaneous change:

Fe(OH)₂(s) + 2e⁻ → Fe(s) + 2OH⁻(aq)      E^θ = −0.88 V

and AgBr(s) + e⁻ → Ag(s) + Br⁻(aq)      E^θ = +0.07 V

Which of the following option is correct?

Consider the above electrochemical cell where a metal electrode ( M ) is undergoing redox reaction by forming $\mathrm{M}^{+}\left(\mathrm{M} \rightarrow \mathrm{M}^{+}+\mathrm{e}^{-}\right)$. The cation $\mathrm{M}^{+}$is present in two different concentrations $c_1$ and $c_2$ as shown above. Which of the following statement is correct for generating a positive cell potential?