The standard reduction potentials of $2H^{+}/H_2$,$Cu^{2+}/Cu$,$Zn^{2+}/Zn$,and $NO_3^{-}, H^{+}/NO$ are $0.0 \ V$,$0.34 \ V$,$-0.76 \ V$,and $0.97 \ V$ respectively. Observe the following reactions:
$I$. $Zn + HCl \rightarrow$
$II$. $Cu + HCl \rightarrow$
$III$. $Cu + HNO_3 \rightarrow$
Which reactions do not liberate $H_{2(g)}$?

Based on the following information,arrange four metals $A, B, C$ and $D$ in order of decreasing ability to act as reducing agents:
$[I]$ Only $A, B$ and $C$ react with $1 \ M \ HCl$ to give $H_2 \ (g)$.
$[II]$ When $C$ is added to solutions of the other metal ions,metallic $B$ and $D$ are formed.
$[III]$ Metal $C$ does not reduce $A^{n+}$.

Which of the following statements is not correct?

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 $3.$

Pick out the incorrect statement $:-$