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:

  • A
    $BrO_3^{-}$
  • B
    $BrO_4^{-}$
  • C
    $Br_2$
  • D
    $HBrO$

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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 $\Lambda_{-}^{\circ} = 50.4 \ S \ cm^{2} \ mol^{-1}$.]

The conductivity of $0.001028 \, mol \, L^{-1}$ acetic acid is $4.95 \times 10^{-5} \, S \, cm^{-1}$. Calculate its dissociation constant if $\Lambda_m^\circ$ for acetic acid is $390.5 \, S \, cm^2 \, mol^{-1}$.

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At $300 \ K$,the $E_{cell}^{\circ}$ of $A_{(s)} + B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + B_{(s)}$ is $1.0 \ V$. If $\Delta_r S^{\circ}$ of this reaction is $100 \ J \ K^{-1} \ mol^{-1}$,what is $\Delta_r H^{\circ}$ (in $kJ \ mol^{-1}$) of this reaction? $(F = 96500 \ C \ mol^{-1})$

The number of correct statements from the following is :
$A.$ $E_{cell}$ is an intensive parameter.
$B.$ $A$ negative $E^{\Theta}$ means that the redox couple is a stronger reducing agent than the $H^{+}/H_2$ couple.
$C.$ The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
$D.$ The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

The value of $E_1^{\circ}$ is (in $V$)

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