Consider the change in oxidation state of Bromine corresponding to different $emf$ values as shown in the diagram below:
$BrO_4^{-}$ $\xrightarrow{1.82 \ V} BrO_3^{-}$ $\xrightarrow{1.5 \ V} HBrO$ $\xrightarrow{1.0652 \ V} Br_2$ $\xrightarrow{1.595 \ V} Br^{-}$
Then the species undergoing disproportionation is:

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:
$(A)$ $24125$ $(B)$ $48250$ $(C)$ $96500$ $(D)$ $193000$

For the reaction $4B_{(s)} + 3O_{2(g)} \rightarrow 2B_2O_{3(g)}$,the standard cell potential is $E^o_{cell} = 1.433 \ V$. Calculate the molar entropy $(S_m^o)$ of oxygen gas in $J/K \ mol$.
Given:
$(\Delta_fH^o)_{B_2O_3(g)} = -840 \ kJ/mol$
$(S_m^o)_{B_2O_3(g)} = 280 \ J/K \ mol$
$(S_m^o)_{B(s)} = 10 \ J/K \ mol$
Assume $\Delta_rG^o = -nFE^o_{cell}$ and $\Delta_rG^o = \Delta_rH^o - T\Delta_rS^o$ at $T = 298 \ K$.

The rusting of iron takes place as follows. Calculate $\Delta G^o$ for the net process in $kJ \ mol^{-1}$.
$2H^{+} + 2e^- + \frac{1}{2}O_2 \longrightarrow H_2O_{(l)} ; E^o = +1.23 \ V$
$Fe^{2+} + 2e^- \longrightarrow Fe_{(s)} ; E^o = -0.44 \ V$

Consider the following electrochemical cell at standard condition: $Au_{(s)} | QH_2, Q | NH_4X(0.01 \ M) || Ag^{+}(1 \ M) | Ag_{(s)}$. Given $E_{\text{cell}} = +0.4 \ V$. The couple $QH_2 / Q$ represents the quinhydrone electrode,and the half-cell reaction is given as: $Q + 2e^- + 2H^+ \rightarrow QH_2$ with $E^o_{Q/QH_2} = +0.7 \ V$. Given: $E^o_{Ag^+/Ag} = +0.8 \ V$ and $\frac{2.303 \ RT}{F} = 0.06 \ V$. The $pK_b$ value of the ammonium halide salt $(NH_4X)$ used here is $.........$ (nearest integer).

On the basis of standard electrode potential values $(E^{\circ})$, suggest which of the following reactions would take place spontaneously? Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$, $E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V$, $E^{\circ}_{Mg^{2+}/Mg} = -2.37 \ V$, $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$, $E^{\circ}_{Br_{2}/Br^{-}} = +1.09 \ V$, $E^{\circ}_{Cl_{2}/Cl^{-}} = +1.36 \ V$, $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$.