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).

  • A
    $5$
  • B
    $6$
  • C
    $16$
  • D
    $9$

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Column $I$Column $II$
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$B$. 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$iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$

For a spontaneous reaction,the $\Delta G$,equilibrium constant $K$,and $E_{Cell}^{o}$ will be respectively:

Which of the following is not correct?

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: $Na=23, Hg=200; 1 \ F = 96500 \ C$).
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$(A)$ $0.5$ $(B)$ $1.0$ $(C)$ $2.0$ $(D)$ $3.0$
$2.$ If the cathode is a $Hg$ electrode,the maximum weight $(g)$ of amalgam formed from this solution is:
$(A)$ $200$ $(B)$ $225$ $(C)$ $400$ $(D)$ $446$
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For a saturated solution of $Ag_{2}CrO_{4}$ at infinite dilution,$\lambda_{m}^{\infty}(Ag^{+}) = 127 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and $\lambda_{m}^{\infty}(CrO_{4}^{2-}) = 246 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$. If the specific conductance of the solution is $2 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$,calculate the solubility product $(K_{sp})$ of $Ag_{2}CrO_{4}$.

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