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Q. No.Calculate emf of the half cell given below :
Pt (s) | \(H_2\) (g, 2 atm) | HCl (aq, 0.02 M)
\(E^0_{H_2 / H^+}\) = 0 V
(Given : \(\frac{2.303 RT}{F} = 0.059\),
log 2 = 0.3010)
1. 0.109 V
2. 0.035 V
3. − 0.035 V
4. − 0.109 V
Pt (s) | \(H_2\) (g, 2 atm) | HCl (aq, 0.02 M)
\(E^0_{H_2 / H^+}\) = 0 V
(Given : \(\frac{2.303 RT}{F} = 0.059\),
log 2 = 0.3010)
1. 0.109 V
2. 0.035 V
3. − 0.035 V
4. − 0.109 V
Subtopic: Nernst Equation |
Level 3: 35%-60%
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A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is:
(Given : Molar mass of Cu = 63 g mol-1; 1 F = 96487 C mol-1)
1. 2.4036 g
2. 1.7018 g
3. 0.5876 g
4. 0.2938 g
(Given : Molar mass of Cu = 63 g mol-1; 1 F = 96487 C mol-1)
1. 2.4036 g
2. 1.7018 g
3. 0.5876 g
4. 0.2938 g
Subtopic: Faraday’s Law of Electrolysis |
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Level 3: 35%-60%
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The standard electrode potential (\(E^°\)) for the half-cell reaction \(Fe^{3+} + e^- \rightarrow Fe^{2+}\) at 298 K is :
(Given: \(E^°(Fe^{3+}/Fe) = -0.04\) V and \(E^°(Fe^{2+}/Fe) = -0.44\) V at 298 K)
(Given: \(E^°(Fe^{3+}/Fe) = -0.04\) V and \(E^°(Fe^{2+}/Fe) = -0.44\) V at 298 K)
| 1. | +0.92 V | 2. | +0.40 V |
| 3. | +0.76 V | 4. | -0.48 V |
Subtopic: Electrode & Electrode Potential |
Level 3: 35%-60%
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For a salt XY, which is a strong electrolyte, the plot of \(\Lambda_m\) versus \(\sqrt{c}\) has a slope of -90.0 S cm² mol-3/2 L1/2 at 298 K. At 0.01 M concentration of XY, the value of \(\Lambda_m\) is 145.0 S cm² mol-1. The limiting molar conductivity of Y- ion (\(\lambda_{Y-}^0\), in S cm² mol-1) at 298 K will be:
(Given: \(\lambda_{X+}^0\) = 74.0 S cm² mol-1)
(Given: \(\lambda_{X+}^0\) = 74.0 S cm² mol-1)
| 1. | 76.0 |
| 2. | 80.0 |
| 3. | 100.0 |
| 4. | 90.0 |
Subtopic: Conductance & Conductivity |
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If the molar conductivity \(\left(\Lambda_{\mathrm{m}}\right)\) of a \(0.050 ~\text{mol}~ \text{L}^{-1}\) solution of a monobasic weak acid is \(90 ~\text{S} ~\text{cm}^2 ~\text{mol}^{-1}\), then its degree of dissociation will be: [Assume \(\Lambda_{+}^0=349.6~ \mathrm{S ~cm}^2 \mathrm{~mol}^{-1}\) and \(\mathrm{\Lambda}_{-}^{\circ}=50.4 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \)]
| 1. | \( 0.225 \) | 2. | \( 0.215 \) |
| 3. | \(0.115 \) | 4. | \(0.125\) |
Subtopic: Conductance & Conductivity |
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NEET - 2025
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Match the redox conversions in List-I with the corresponding number of Faradays required in List-II.
Choose the correct answer from the options given below:
| List-I (Redox Conversion) |
List-II (Number of Faraday required) |
||
| A. | 1 mol of H2O to O2 | I. | 3F |
| B. | 1 mol of \(MnO^-_4\) to \(Mn^{2+}\) | II. | 2F |
| C. | 1.5 mol of \(Ca\) from molten \(CaCl_2\) | III. | 1F |
| D. | 1 mol of FeO to Fe2O3 | IV. | 5F |
Choose the correct answer from the options given below:
| 1. | A - III, B - IV, C - I, D - II | 2. | A - II, B - III, C - I, D - IV |
| 3. | A - III, B - IV, C - II, D - I | 4. | A - II, B - IV, C - I, D - III |
Subtopic: Faraday’s Law of Electrolysis |
65%
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Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate for 100 seconds is:
(Given : Molar mass of Cu: \(63~ g mol^{-1}\); 1F = 96487 C)
(Given : Molar mass of Cu: \(63~ g mol^{-1}\); 1F = 96487 C)
| 1. | 0.315 g | 2. | 31.5 g |
| 3. | 0.0315 g | 4. | 3.15 g |
Subtopic: Faraday’s Law of Electrolysis |
68%
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Given below are two statements:
In the light of the above statements, choose the correct answer from the options given below:
1. Both Statement I and Statement II are true
2. Both Statement I and Statement II are false
3. Statement I is true but Statement II is false
4. Statement I is false but Statement II is true
| Statement I: | \(2 \mathrm{~F}\) electricity is required for the oxidation of 1 mole \(\mathrm{H}_2 \mathrm{O}\) to \(\mathrm{O}_2\). |
| Statement II: | To get \(40.0 \mathrm{~g}\) of Aluminium from molten \(\mathrm{Al}_2 \mathrm{O}_3\) required electricity is \(4.44 \mathrm{~F}\). |
In the light of the above statements, choose the correct answer from the options given below:
1. Both Statement I and Statement II are true
2. Both Statement I and Statement II are false
3. Statement I is true but Statement II is false
4. Statement I is false but Statement II is true
Subtopic: Faraday’s Law of Electrolysis |
62%
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From the following select the one which is not an example of corrosion:
| 1. | Rusting of an iron object |
| 2. | Production of hydrogen by electrolysis of water |
| 3. | Tarnishing of silver |
| 4. | Development of green coating on copper and bronze ornaments |
Subtopic: Corrosion |
83%
Level 1: 80%+
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The standard cell potential of the following cell \(\mathrm{{Zn}\left|{Zn}^{2+}({aq}) \| {Fe}^{2+}({aq})\right| {Fe}~\text{ is }~0.32 {~V}}.\) Calculate the standard Gibbs energy change for the reaction:
\(\mathrm{{Zn}({s})+{Fe}^{2+}({aq}) \rightarrow {Zn}^{2+}({aq})+{Fe}({s})}\)
(Given : \(1 \mathrm{~F}=96487 \mathrm{C} mol^{-1}\))
\(\mathrm{{Zn}({s})+{Fe}^{2+}({aq}) \rightarrow {Zn}^{2+}({aq})+{Fe}({s})}\)
(Given : \(1 \mathrm{~F}=96487 \mathrm{C} mol^{-1}\))
| 1. | \(-61.75 \mathrm{{~kJ} {~mol}}^{-1}\) | 2. | \(+5.006 \mathrm{{~kJ} {~mol}}^{-1}\) |
| 3. | \(-5.006 \mathrm{{~kJ} {~mol}}^{-1}\) | 4. | \(+61.75 \mathrm{{~kJ} {~mol}}^{-1}\) |
Subtopic: Relation between Emf, G, Kc & pH |
84%
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