What are the uses of standard half-cell potential?

Consider the cell reaction at $300 \ K$: $A_{(s)} + B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + B_{(s)}$. Its $E^{\circ}$ is $1.0 \ V$. The $\Delta_{r}H^{\circ}$ of the reaction is $-163 \ kJ \ mol^{-1}$. What is $\Delta_{r}S^{\circ}$ (in $J \ K^{-1} \ mol^{-1}$) of the reaction? $(F = 96500 \ C \ mol^{-1})$

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

Which reactions are possible due to electro-chemistry in human brain?

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential $(E^{\circ})$ of two half-cell reactions decide which way the reaction is expected to proceed. $A$ simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}$ ($V$ with respect to normal hydrogen electrode) values.
$I_2 + 2e^{-} \rightarrow 2I^{-} \quad E^{\circ} = 0.54 \ V$
$Cl_2 + 2e^{-} \rightarrow 2Cl^{-} \quad E^{\circ} = 1.36 \ V$
$Mn^{3+} + e^{-} \rightarrow Mn^{2+} \quad E^{\circ} = 1.50 \ V$
$Fe^{3+} + e^{-} \rightarrow Fe^{2+} \quad E^{\circ} = 0.77 \ V$
$O_2 + 4H^{+} + 4e^{-} \rightarrow 2H_2O \quad E^{\circ} = 1.23 \ V$
$1.$ Among the following,identify the correct statement.
$(A)$ Chloride ion is oxidized by $O_2$
$(B)$ $Fe^{2+}$ is oxidized by iodine
$(C)$ Iodide ion is oxidized by chlorine
$(D)$ $Mn^{2+}$ is oxidized by chlorine
$2.$ While $Fe^{3+}$ is stable,$Mn^{3+}$ is not stable in acid solution because
$(A)$ $O_2$ oxidizes $Mn^{2+}$ to $Mn^{3+}$
$(B)$ $O_2$ oxidizes both $Mn^{2+}$ and $Fe^{2+}$ to $Fe^{3+}$
$(C)$ $Fe^{3+}$ oxidizes $H_2O$ to $O_2$
$(D)$ $Mn^{3+}$ oxidizes $H_2O$ to $O_2$
$3.$ Sodium fusion extract,obtained from aniline,on treatment with iron$(II)$ sulphate and $H_2SO_4$ in presence of air gives a Prussian blue precipitate. The blue color is due to the formation of
$(A)$ $Fe_4[Fe(CN)_6]_3$
$(B)$ $Fe_3[Fe(CN)_6]_2$
$(C)$ $Fe_4[Fe(CN)_6]_2$
$(D)$ $Fe_3[Fe(CN)_6]_3$
Give the answer for questions $1, 2$ and $3.$

For a spontaneous reaction,determine the values of $\Delta G^o$,equilibrium constant $K$,and $E^o_{cell}$ respectively.