Based on the following information,arrange four metals,$A$,$B$,$C$,and $D$,in order of increasing ability to act as reducing agents:
$(I)$ Only $C$ reacts with $1 \ M \ HCl$ to give $H_{2(g)}$.
$(II)$ When $A$ is added to a solution of the other metal salts,metallic $D$ is formed,but not $B$ or $C$.

Match the column $I$ with column $II$ and mark the appropriate choice.

Column $I$	Column $II$
$A$. Kohlrausch law	$i$. $\Lambda _{m}^o = \nu _+ \lambda _+^o + \nu _- \lambda _-^o$
$B$. Molar Conductivity	$ii$. $\Lambda _m = \frac{\kappa \times 1000}{M}$
$C$. Degree of Dissociation	$iii$. $\alpha = \frac{\Lambda _m}{\Lambda _m^o}$
$D$. Dissociation Constant	$iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$

Consider a $70 \%$ efficient hydrogen-oxygen fuel cell working under standard conditions at $1 \ bar$ and $298 \ K$. Its cell reaction is
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O(\ell)$
The work derived from the cell on the consumption of $1.0 \times 10^{-3} \ mol$ of $H_{2(g)}$ is used to compress $1.00 \ mol$ of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in $K$) of the ideal gas?
The standard reduction potentials for the two half-cells are given below.
$O_{2(g)} + 4H^{+}(aq.) + 4e^- \rightarrow 2H_2O(\ell), E^{\circ} = 1.23 \ V$
$2H^{+}(aq.) + 2e^- \rightarrow H_{2(g)}, E^{\circ} = 0.00 \ V$
Use $F = 96500 \ C \ mol^{-1}, R = 8.314 \ J \ mol^{-1} \ K^{-1}$

If the molar conductivity $(\Lambda_{m})$ of a $0.050 \ mol \ L^{-1}$ solution of a monobasic weak acid is $90 \ S \ cm^{2} \ mol^{-1}$,its extent (degree) of dissociation will be. [Assume $\Lambda_{+}^{\circ} = 349.6 \ S \ cm^{2} \ mol^{-1}$ and $\Lambda_{-}^{\circ} = 50.4 \ S \ cm^{2} \ mol^{-1}$.]

Given:
$\lambda _{ClCH_2COONa} = 224 \ \Omega ^{-1} \ cm^2 \ gmeq^{-1}$
$\lambda _{NaCl} = 38.2 \ \Omega ^{-1} \ cm^2 \ gmeq^{-1}$
$\lambda _{HCl} = 203 \ \Omega ^{-1} \ cm^2 \ gmeq^{-1}$
What is the value of $\lambda _{ClCH_2COOH}$ in $\Omega ^{-1} \ cm^2 \ gmeq^{-1}$?

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 $3.$