During the electrolysis of concentrated $H_2SO_4$,perdisulphuric acid $(H_2S_2O_8)$ and $O_2$ are formed at the anode in equimolar amounts. The moles of $H_2$ that will form simultaneously at the other electrode will be (Given: $2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^-$)
- Athrice of $O_2$
- Btwice of $O_2$
- Cequal to $O_2$
- Dhalf of $O_2$
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All the energy released from the reaction $X \rightarrow Y, \Delta_{r}G^0 = -193 \ kJ \ mol^{-1}$ is used for oxidizing $M^{+}$ as $M^{+} \rightarrow M^{3+} + 2e^-, E^0 = -0.25 \ V$. Under standard conditions,the number of moles of $M^{+}$ oxidized when one mole of $X$ is converted to $Y$ is $\left[F = 96500 \ C \ mol^{-1}\right]$.
MediumIIT 2015
View SolutionMatch List-$I$ with List-$II$.
List-$I$ List-$II$ $A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$ $I$. Primary battery $B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$ $II$. Discharging of secondary battery $C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$ $III$. Fuel cell $D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$ $IV$. Charging of secondary battery
Choose the correct answer from the options given below.
| List-$I$ | List-$II$ |
| $A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$ | $I$. Primary battery |
| $B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$ | $II$. Discharging of secondary battery |
| $C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$ | $III$. Fuel cell |
| $D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$ | $IV$. Charging of secondary battery |
Choose the correct answer from the options given below.
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 $iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$
| 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 | $iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$ |
Medium
View SolutionFind the solubility product $(K_{sp})$ of a saturated solution of $Ag_2CrO_4$ in water at $298 \ K$,if the $emf$ of the cell $Ag | Ag^{+} (\text{satd. } Ag_2CrO_4 \text{ solution}) || Ag^{+} (0.1 \ M) | Ag$ is $0.591 \ V$ at $298 \ K$.
Medium
View SolutionMatch List-$I$ with List-$II$.
List-$I$ (Conversion) List-$II$ (Number of Faraday required) $A$. $1 \ mol$ of $H_2O$ to $O_2$ $I$. $3 \ F$ $B$. $1 \ mol$ of $MnO_4^{-}$ to $Mn^{2+}$ $II$. $2 \ F$ $C$. $1.5 \ mol$ of $Ca$ from molten $CaCl_2$ $III$. $1 \ F$ $D$. $1 \ mol$ of $FeO$ to $Fe_2O_3$ $IV$. $5 \ F$
Choose the correct answer from the options given below:
| List-$I$ (Conversion) | List-$II$ (Number of Faraday required) |
| $A$. $1 \ mol$ of $H_2O$ to $O_2$ | $I$. $3 \ F$ |
| $B$. $1 \ mol$ of $MnO_4^{-}$ to $Mn^{2+}$ | $II$. $2 \ F$ |
| $C$. $1.5 \ mol$ of $Ca$ from molten $CaCl_2$ | $III$. $1 \ F$ |
| $D$. $1 \ mol$ of $FeO$ to $Fe_2O_3$ | $IV$. $5 \ F$ |
Choose the correct answer from the options given below:
MediumNEET 2024
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