A button cell used in watches functions as following

Zn(s) + Ag₂O(s) + H₂O (l) $\rightleftharpoons$2Ag(s) + Zn²⁺(aq) + 2OH^-(aq)

If half cell potentials are

$$\begin{gathered} {{\mathrm{Zn}}^{2+}}_{\left(\mathrm{aq}\right)} +\hspace{0.17em}2{\mathrm{e}}^{-} \to {\mathrm{Zn}}_{\left(\mathrm{x}\right)}; {\mathrm{E}}^{\mathrm{o}} = -0.76\mathrm{V} \\ {\mathrm{Ag}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)} + {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)} + 2{\mathrm{e}}^{-} \to 2{\mathrm{Ag}}_{\left(\mathrm{s}\right)} + 2{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)}; \\ {\mathrm{E}}^{\mathrm{o}} = 0.34\mathrm{V} \end{gathered}$$

The cell potential will be

1. 0.84 V

2. 1.34 V

3. 1.10 V

4. 0.42 V

 

 

Limiting molar conductivity of NH₄OH (i.e Åm(NH₄OH)) is equal to:-

1. Å_m (NH₄Cl)+Å_m(NaCl)-Å_m(NaOH) 

2. Å_m(NaOH)+Å_m(NaCl)-Å_m(NH₄Cl)

3. Å_m(NH₄OH)+Å_m(NH₄Cl)-Å_m(HCl)

4. Å(NH₄Cl)+Å(NaOH)-Å(NaCl)

Kohlrausch's law states that at

(1) finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.

(2) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of th electrolyte.

(3) infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte.

(4) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.

Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4 and -8.2 for H₂O(l), CO₂(g) and pentane (g), respectively. The value of E°_cell for the pentane-oxygen fuel cell is

(1) 1.968 V

(2) 2.0968 V

(3) 1.0968 V

(4) 0.0968 V

The equilibrium constant of the reaction:

Cu (s) + 2Ag⁺ (aq) $\to$Cu²⁺ (aq) + 2Ag(s) ;

E^o = 0.46 V at 298 K is:

The efficiency of a fuel cell is given by:

$\left(\mathrm{a}\right) ∆\mathrm{H}/∆\mathrm{G} \left(\mathrm{b}\right) ∆\mathrm{G}/∆\mathrm{S}$
$\left(\mathrm{c}\right) ∆\mathrm{G}/∆\mathrm{H} \left(\mathrm{d}\right) ∆\mathrm{S}/∆\mathrm{G}$

A hypothetical electrochemical cell is shown below A l A⁺ (xM)l l B⁺ (yM) l B
The Emf measured is +0.20 V.The cell reaction is:-
(1) A⁺ + B → A + B⁺

(2) A⁺ + e^- →  A ; B⁺ + e^- → B

(3) the cell reaction cannot be predicted

(4) A + B⁺ → A⁺ + B

The E° in the given diagram is,

(1) 0.5

(2) 0.6

(3) 0.7

(4) 0.8

At 298K the standard free energy of formation of H₂O(l) is –237.20kJ/mole while that of its ionisation into H+ iion and hydroxyl ions is 80 kJ/mole, then the emf of the following cell at 298 K will be

H₂(g,1 bar) | H⁺ (1M) || OH^–(1M) | O₂ (g, 1bar)

(1) 0.40 V

(2) 0.81 V

(3) 1.23 V

(4) –0.40 V

Which of the following cell can produce more electric work.

(1) pt,H₂|NH₄Cl||0.1MCH₃COOH|H₂,pt

(2) pt,H₂|0.1MHCl||0.1MNaOH|H₂,pt

(3) pt,H₂|0.1MHCl||0.1MCH₃COOK|H₂,pt

(4) pt,H₂|0.1MCH₃COOK||0.1MHCl|H₂,pt