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
2. velocity of is greater than that of K+
3. Velocities of both K+ and 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
The mass of silver(eq. mass = 108) displaced by that quantity of current which displaced 5600 mL of hydrogen at STP is:
1. 54 g
2. 108 g
3. 5.4 g
4. none of these
A silver cup is plated with silver by passing 965 coulomb of electricity. The amount of Ag deposited is:
1. 1.08 g
2. 1.0002 g
3. 9.89 g
4. 107.89 g
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:
1. | 0.142 cm–1 | 2. | 0.66 cm–1 |
3. | 0.918 cm–1 | 4. | 1.12 cm–1 |
Given below are two half-cell reactions:
Mn2+ + 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.