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 $(-2.87 \mathrm{~V})$.
Assertion (A) Cu is less reactive than hydrogen.
Reason $(R) \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{S}}$ is negative.
Assertion (A) $\mathrm{E}_{\text {cell }}$ should have a positive value for the cell to function.
Reason (R) $\mathrm{E}_{\text {cathode }}<\mathrm{E}_{\text {anode }}$