Consider the following electrodes
$$\begin{aligned} & P=\mathrm{Zn}^{2+}(0.0001 \mathrm{M}) / \mathrm{Zn}, Q=\mathrm{Zn}^{2+}(0.1 \mathrm{M}) / \mathrm{Zn} \\ & R=\mathrm{Zn}^{2+}(0.01 \mathrm{M}) / \mathrm{Zn}, S=\mathrm{Zn}^{2+}(0.001 \mathrm{M}) / \mathrm{Zn} \end{aligned}$$
$$E^{\circ}(\mathrm{Zn} / \mathrm{Zn}^{2+})=-0.76 \mathrm{~V}$$ electrode potentials of the above electrodes in volts are in the order
The number of angular and radial nodes in $$3 p$$ orbital respectively are
The resistance of $$0.01 \mathrm{~m} \mathrm{~KCl}$$ solution at $$298 \mathrm{~K}$$ is $$1500 \Omega$$. If the conductivity of $$0.01 \mathrm{~m} \mathrm{~KCl}$$ solution at $$298 \mathrm{~K}$$ is $$0.1466 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1}$$. The cell constant of the conductivity cell in $$\mathrm{cm}^{-1}$$ is
$$\mathrm{H}_2(g)+2 \mathrm{AgCl}(s) \rightleftharpoons 2 \mathrm{Ag}(s)+2 \mathrm{HCl}(a q)$$
$$E_{\text {cell }}^{\circ}$$ at $$25^{\circ} \mathrm{C}$$ for the cell is $$0.22 \mathrm{~V}$$. The equilibrium constant at $$25^{\circ} \mathrm{C}$$ is