An electrochemical cell is shown below Pt, H₂(1 atm)| HCI (0.1 M)CH₃COOH (0.1 M)|

H₂(1 atm), Pt The EMF of the cell will not be zero, because

1. EMF depends on molarities of acids used

2. pH of 0.1 M HCl and 0.1 M CH₃COOH is not same

3. the temperature is constant

4. acids used in two compartments are different

Saturated solution of KNO₃ is used to make 'salt-bridge' because:

1. velocity of K⁺ is greater than that of ${\mathrm{NO}}_3^{-}$

2. velocity of ${\mathrm{NO}}_3^{-}$ is greater than that of K⁺

3. Velocities of both K⁺ and ${\mathrm{NO}}_3^{-}$ are nearly the same

4. KNO₃ is highly soluble in water

A current is passed through two voltameters connected in series. The first voltmeter connected in series. The first voltmeter contains XSO₄(aq) while the second voltmeter contains Y₂SO₄(aq). The relative  atomic masses of X and Y are in the ratio of 2:1. The ratio of the mass of X liberated to the mass of Y liberated is: 

1. 1:1

2. 1:2

3. 2:1

4. None of the above

At 25 °C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 Ω^-1 cm² mol^-1 and at infinite dilution its molar conductance is 238 Ω^-1 cm² mol^-1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is :

Which is the correct representation for Nernst equation ?

1. \(E_{\mathrm{RP}}=E_{\mathrm{RP}}^{\circ}+\frac{0.059}{\mathrm{n}} \log \frac{[\text { oxidant }]}{[\text { reductant }]}\)

2. \(E_{\mathrm{OP}}=E_{\mathrm{OP}}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{[\text { oxidant }]}{[\text { reductant }]}\)

3. \(E_{\mathrm{OP}}=E_{\mathrm{OP}}^{\circ}+\frac{0.059}{\mathrm{n}} \log \frac{[\text { reductant }]}{[\text { oxidant }]}\)

4. All of the above

When a copper wire is immersed in a solution of AgNO₃, the colour of the solution becomes blue because copper:

1. Forms a soluble complex with \(AgNO_3\)
2. Is oxidised to \(Cu^{2+}\)
3. Is reduced to \(Cu^{2-}\)
4. Splits up into atomic form and dissolves

The specific conductance of a 0.1 M KCl solution at 23 °C is 0.012  Ω^–1 cm^–1. The resistance of cells containing the solution at the same temperature was found to be 55 Ω. The cell constant will be:

Given below are two half-cell reactions:

Mn^₂₊ + 2e^-  → Mn;          E₀ = -1.18V

2Mn³⁺ + 2e^-  → 2Mn²⁺;  E₀ = +1.51V

The E₀ for 3Mn²⁺ → 2Mn⁺³  +  Mn  will be:

1. -2.69V; the reaction will not occur

2. -2.69V; the reaction will occur

3. -0.33V; the reaction will not occur

4. -0.33V; the reaction will occur

E⁰ for Fe²⁺ + 2e → Fe is –0.44 volt and E⁰ for Zn²⁺ + 2e→ Zn is –0.76 volt, thus:

1. Zn is more electropositive than Fe.

2. Fe is more electropositive than Zn.

3. Zn is more electronegative.

4. None of the above.

The voltage of the cell given below increases with: 

Cell: Sn(s) + 2Ag⁺(aq) → Sn²⁺(aq) + 2Ag(s)

1. Increase in size of the silver rod.

2. Increase in the concentration of Sn²⁺ ions.

3. Increase in the concentration of Ag⁺ ions.

4. None of the above.