Match the terms given in Column I with the items given in Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | $\wedge_m$ | 1. | Intensive property |
| B. | $E^s_{\text{cell}}$ | 2. | Depends on number of ions/volume |
| C. | $\kappa$ | 3. | Extensive property |
| D. | $\Delta_r G_{\text{cell}}$ | 4. | Increases with dilution |
A. $\rightarrow$ (4) B. $\rightarrow$ (1) C. $\rightarrow$ (2) D. $\rightarrow$ (3)
A. $\wedge_m$ (molar conductivity) is the conductivity due to number of ions furnished by one mole of electrolyte. As dilution increases number of ions present in the solution increases hence molar conductivity increases.
B. $E_{\text {cell }}^{\circ}$ of any atom/ion does not depend upon number of atom/ion, hence $E_{\text {cell }}^{\circ}$ of any atom/ion is an intensive properties.
C. $\kappa$ represents specific conductivity which depends upon number of ions present in per unit volume.
D. $\Delta_r G_{\text {cell }}$ is an extensive property as it depends upon number of particles(species).
Match the items of Column I and Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | Lead storage battery | 1. | Maximum efficiency |
| B. | Mercury cell | 2. | Prevented by galvanisation |
| C. | Fuel cell | 3. | Gives steady potential |
| D. | Rusting | 4. | Pb is anode, PbO$_2$ is cathode |
A. $\rightarrow(4)$ B. $\rightarrow$ (3) C. $\rightarrow$ (1) D. $\rightarrow$ (2)
A. Chemical reaction occurring on lead storage battery can be represented as
At anode $\mathrm{Pb}(\mathrm{s})+\mathrm{SO}_4{ }^{2-}(\mathrm{aq}) \longrightarrow \mathrm{PbSO}_4(\mathrm{~s})+2 e^{-}$
At cathode $\mathrm{PbO}_2(\mathrm{~s})+\mathrm{SO}_4^{2-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq}) \xrightarrow{+2 e^{-}} 2 \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(l)$
Thus, Pb is anode and $\mathrm{PbO}_2$ is cathode.
B. Mercury cell does not include ions during their function hence produce steady current.
C. Fuel cell has maximum efficiency as they produce energy due to combustion reaction of fuel.
D. Rusting is prevented by corrosion.
Match the items of Column I and Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | $\kappa$ | 1. | $I \times t$ |
| B. | $\wedge_m$ | 2. | $\wedge_m / \wedge^o_m$ |
| C. | $\alpha$ | 3. | $\frac{\kappa}{C}$ |
| D. | $Q$ | 4. | $\frac{G^*}{R}$ |
A. $\rightarrow(4)$ B. $\rightarrow$ (3) C. $\rightarrow(2)$ D. $\rightarrow$ (1)
A. Conductivity $(\kappa)=\frac{G^*}{R}$
B. Molar conductivity $\left(\wedge_m\right)=\frac{\kappa}{C}$
C. Degree of dissociation $(\alpha)=\frac{\wedge_m}{\wedge_m^{\circ}}$
D. Charge $Q=I \times t$
where, $Q$ is the quantity of charge in coulomb when $I$ ampere of current is passed through an electrolyte for $t$ second.
Match the items of Column I and Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | Lechlanche cell | 1. | Cell reaction $2 \mathrm{H}_2+\mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}$ |
| B. | Ni-Cd cell | 2. | Does not involve any ion in solution and is used in hearing aids. |
| C. | Fuel cell | 3. | Rechargeable |
| D. | Mercury cell | 4. | Reaction at anode, $\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 e^{-}$ |
| 5. | Converts energy of combustion into electrical energy |
A. $\rightarrow$ (4) B). $\rightarrow$ (3) C. $\rightarrow(1,5)$ D. $\rightarrow$ (2)
A. Lechlanche cell The electrode reaction occurs at Lechlanche cell are
At anode $\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Zn}^2+2 \mathrm{e}^{-}$
At cathode $\mathrm{MnO}_2+\mathrm{NH}_4^{+}+e^{-} \longrightarrow \mathrm{MnO}(\mathrm{OH})+\mathrm{NH}_3$
B. Ni-Cd cell is rechargeable. So, it has more life time.
C. Fuel cell produces energy due to combustion. So, fuel cell converts energy of combustion into electrical energy e.g., $2 \mathrm{H}_2+\mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}$
D. Mercury cell does not involve any ion in solution and is used in hearing aids.
Match the items of Column I and Column II on the basis of data given below
$$\begin{aligned} \mathrm{E}_{\mathrm{F}_2 / \mathrm{F}^{-}}^{\mathrm{s}} & =2.87 \mathrm{~V}, \mathrm{E}_{\mathrm{Li}^{+} / \mathrm{Li}}^{\mathrm{s}}=-3.5 \mathrm{~V}, \\ \mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}} & =1.4 \mathrm{~V}, \mathrm{E}_{\mathrm{Br}_2 / \mathrm{Br}^{-}}^{\mathrm{s}}=1.09 \mathrm{~V} \end{aligned}$$
| Column I | Column II | ||
|---|---|---|---|
| A. | F$_2$ | 1. | Metal is the strongest reducing agent |
| B. | Li | 2. | Metal ion which is the weakest oxidising agent |
| C. | Au$^{3+}$ | 3. | Non-metal which is the best oxidising agent |
| D. | Br$^{-}$ | 4. | Unreactive metal |
| E. | Au | 5. | Anion that can be oxidised by Au$^{3+}$ |
| F. | Li$^{+}$ | 6. | Anion which is the weakest reducing agent |
| G. | F$^-$ | 7. | Metal ion which is an oxidising agent |
A. $\rightarrow(3)$ B. $\rightarrow$ (1) C. $\rightarrow$ (7) D. $\rightarrow$ (5) E. $\rightarrow$ (4) F. $\rightarrow$ (2) G. $\rightarrow(6)$
A. $F_2$ is a non-metal and best oxidising agent because SRP of $F_2$ is +2.87 V .
B. Li is a metal and strongest reducing agent because SRP of Li is -3.05 V .
C. $\mathrm{Au}^{3+}$ is a metal ion which is an oxidising agent as SRP of $\mathrm{Au}^{3+}$ is +1.40 V .
D. $\mathrm{Br}^{-}$is an anion that can be oxidised by $\mathrm{Au}^{3+}$ as $\mathrm{Au}^{3+}\left(E^{\circ}=1.40\right)$ is greater than $$\mathrm{Br}^{-}\left(E^{\circ}=1.09 \mathrm{~V}\right)$$
E . Au is an unreactive metal.
F. $\mathrm{Li}^{+}$is a metal ion having least value of $\operatorname{SRP}(-3.05 \mathrm{~V})$, hence it is the weakest oxidising agent.
G. $\mathrm{F}^{-}$ is an anion which is the weakest reducing agent as $\mathrm{F}^{-} / \mathrm{F}_2$ has low oxidation potential $(-2.87 \mathrm{~V})$.