For the fuel cell reaction:
$2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}$; $\Delta_fH^o_{298}(H_2O_{(l)}) = -285.5 \ kJ/mol$
What is $\Delta S^o_{298}$ for the given fuel cell reaction?
Given: $O_{2(g)} + 4H^+_{(aq)} + 4e^- \to 2H_2O_{(l)}$; $E^o = 1.23 \ V$

On the basis of standard electrode potential values,suggest which of the following reactions would take place?

During electrolysis of aqueous $NaOH$,$4 \ g$ of $O_2$ gas is liberated at $NTP$ at the anode. The volume of $H_2$ gas liberated at the cathode in $litres$ is $..............$

The reduction potentials of four elements $P, Q, R,$ and $S$ are $-2.90 \ V, +0.34 \ V, +1.20 \ V,$ and $-0.76 \ V$ respectively. The order of decreasing reactivity is:

At $298 \ K$,the equilibrium constant is $2 \times 10^{15}$ for the reaction:
$Cu_{(s)} + 2 Ag^{+}_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2 Ag_{(s)}$
The equilibrium constant for the reaction $\frac{1}{2} Cu^{2+}_{(aq)} + Ag_{(s)} \rightleftharpoons \frac{1}{2} Cu_{(s)} + Ag^{+}_{(aq)}$ is $x \times 10^{-8}$. The value of $x$ is (Nearest Integer).

The equilibrium constant of a $2$ electron redox reaction at $298 \, K$ is $3.8 \times 10^{-3}$. The cell potential $E^{\circ}$ (in $V$) and the free energy change $\Delta G^{\circ}$ (in $kJ \, mol^{-1}$) for this equilibrium,respectively are