Given below are the half-cell reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn; E^{o} = -1.18 \ V$
$2(Mn^{3+} + e^{-} \rightarrow Mn^{2+}); E^{o} = +1.51 \ V$
The $E^{o}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be:

Plotting $1 / \Lambda_{m}$ against $c \Lambda_{m}$ for aqueous solutions of a monobasic weak acid $(HX)$ resulted in a straight line with $y$-axis intercept of $P$ and slope of $S$. The ratio $P / S$ is
$[\Lambda_{m} =$ molar conductivity
$\Lambda_{m}^{\circ} =$ limiting molar conductivity
$c =$ molar concentration
$K_{a} =$ dissociation constant of $HX$ ]

Consider an electrochemical cell: $A_{(s)} | A^{n+}(aq, 2 \ M) || B^{2n+}(aq, 1 \ M) | B_{(s)}$. The value of $\Delta H^{\ominus}$ for the cell reaction is twice that of $\Delta G^{\ominus}$ at $300 \ K$. If the emf of the cell is zero,the $\Delta S^{\ominus}$ (in $J \ K^{-1} \ mol^{-1}$) of the cell reaction per mole of $B$ formed at $300 \ K$ is. . . . . . . (Given: $\ln(2) = 0.7, R = 8.3 \ J \ K^{-1} \ mol^{-1}$.)

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 $3.$

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 $H_{2(g)}$?

Chemical reactions involve the interaction of atoms and molecules. $A$ large number of atoms/molecules (approximately $6.023 \times 10^{23}$) are present in a few grams of any chemical compound,varying with their atomic/molecular masses. To handle such large numbers conveniently,the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry,biochemistry,electrochemistry,and radiochemistry. The following example illustrates a typical case involving a chemical/electrochemical reaction,which requires a clear understanding of the mole concept. $A$ $4.0 \ M$ aqueous solution of $NaCl$ is prepared and $500 \ mL$ of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: $Na=23, Hg=200; 1 \ F = 96500 \ C$).
$1.$ The total number of moles of chlorine gas evolved is:
$(A)$ $0.5$ $(B)$ $1.0$ $(C)$ $2.0$ $(D)$ $3.0$
$2.$ If the cathode is a $Hg$ electrode,the maximum weight $(g)$ of amalgam formed from this solution is:
$(A)$ $200$ $(B)$ $225$ $(C)$ $400$ $(D)$ $446$
$3.$ The total charge (coulombs) required for complete electrolysis is:
$(A)$ $24125$ $(B)$ $48250$ $(C)$ $96500$ $(D)$ $193000$