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Standard electrode potential of three metals X, Y and Z are $-1.2 \text{ V}$, $+0.5 \text{ V}$ and $-3.0 \text{ V}$ respectively. The reducing power of these metals will be:

Which of the following cannot act both as a Bronsted acid and as a Bronsted base?

Which of the following molecular hydrides acts as a Lewis acid?

The EMF of a Daniel cell at $298\text{ K}$ is $E_1$: $Zn|ZnSO_4(0.01\text{ M}) || CuSO_4(1.0\text{ M})|Cu$. When the concentration of $ZnSO_4$ is $1.0\text{ M}$ and that of $CuSO_4$ is $0.01\text{ M}$, the EMF is changed to $E_2$. The correct relationship between $E_1$ and $E_2$ is:

Limiting molar conductivity of $\text{NH}_4\text{OH}$ (i.e., $\Lambda^\circ_m(\text{NH}_4\text{OH})$) is equal to:

The correct relation between dissociation constants of a di-basic acid is:

The compound with the highest pH among the following is:

Limiting molar conductivity of $\text{NH}_4\text{OH}$ (i.e. $\Lambda^\circ_m(\text{NH}_4\text{OH})$) is equal to:

Given below are half-cell reactions: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \ ; \ E^\circ_{Mn^{2+}/MnO_4^-} = -1.510\text{ V}$ $\frac{1}{2}O_2 + 2H^+ + 2e^- \rightarrow H_2O \ ; \ E^\circ_{O_2/H_2O} = +1.223\text{ V}$ Will the permanganate ion, $MnO_4^-$, liberate $O_2$ from water in the presence of an acid?

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below. Then the species undergoing disproportionation is:

Consider the following relations for emf of an electrochemical cell: (a) emf of a cell = (Oxidation potential of the anode) – (Reduction potential of the cathode) (b) emf of a cell = (Oxidation potential of the anode) + (Reduction potential of the cathode) (c) emf of a cell = (Reduction potential of the anode) + (Reduction potential of the cathode) (d) emf of a cell = (Oxidation potential of the anode) – (Oxidation potential of the cathode) Which of the following combinations correctly represents the relation for the emf of the cell?

A hypothetical electrochemical cell is shown below: $\text{A} | \text{A}^+(x\text{ M}) || \text{B}^+(y\text{ M}) | \text{B}$ The EMF measured is $+0.20 \text{ V}$. The cell reaction is:

The solubility of AgCl(s) with solubility product $1.6 \times 10^{-10}$ in 0.1 M NaCl solution would be:

The ionisation constant of ammonium hydroxide is $1.77 \times 10^{-5}$ at 298 K. Hydrolysis constant of ammonium chloride is:

Consider the following reaction: $\frac{4}{3}Al(s) + O_2(g) \rightarrow \frac{2}{3}Al_2O_3(s) \ ; \ \Delta G = -827 \text{ kJ mol}^{-1}$ The minimum e.m.f. required to carry out the electrolysis of $Al_2O_3$ is: (Given $F = 96500 \text{ C mol}^{-1}$)

A button cell used in watches functions as following: $\text{Zn}(s) + \text{Ag}_2\text{O}(s) + \text{H}_2\text{O}(l) \rightleftharpoons 2\text{Ag}(s) + \text{Zn}^{2+}(aq) + 2\text{OH}^-(aq)$ If half-cell potentials are: $\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad E^\circ = -0.76 \text{ V}$ $\text{Ag}_2\text{O}(s) + \text{H}_2\text{O}(l) + 2e^- \rightarrow 2\text{Ag}(s) + 2\text{OH}^-(aq) \quad E^\circ = 0.34 \text{ V}$ The cell potential will be:

In which of the following compounds, the C—Cl bond ionization shall give the most stable carbonium ion?

For a given exothermic reaction, $K_p$ and $K'_p$ are the equilibrium constants at temperatures $T_1$ and $T_2$ respectively. Assuming that heat of reaction is constant in a temperature range between $T_1$ and $T_2$, it is readily observed that (assuming $T_2 > T_1$):

Standard free energies of formation (in $\text{kJ/mol}$) at $298 \text{ K}$ are $-237.2$, $-394.4$ and $-8.2$ for $\text{H}_2\text{O}(l)$, $\text{CO}_2(g)$ and pentane (g), respectively. The value of $E^\circ_{\text{cell}}$ for the pentane-oxygen fuel cell is

Given below are two statements: Statement I: $2\text{ F}$ electricity is required for the oxidation of $1\text{ mole H}_2\text{O}$ to $\text{O}_2$. Statement II: To get $40.0\text{ g}$ of Aluminium from molten $\text{Al}_2\text{O}_3$ required electricity is $4.44\text{ F}$. In the light of the above statements, choose the correct answer from the options given below: