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NEET NEET Questions — Free MCQ Practice

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Value of ${Λ_{m}}^{\infty}$ for SrCl₂ in water at 25°C from the following data:

Conc. (mol/lt) 0.25 1

$Λ_{m}$ (Ω^–1 cm² mol^–1) 260 250

A.

270

B.

260

C.

250

D.

255

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Explanation

$Λ_{m} = Λ_{m}^{\infty} - b \sqrt{c}$

260 = $Λ_{m}^{\infty}$ – 0.5 b…

250 = $Λ_{m}^{\infty}$ – b … (2)

On solving & (2), we get

$Λ_{m}^{\infty}$ = 270 Ω^– cm²mol^–1

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Calculate the useful work of the reaction Ag(s) + 1/2Cl₂(g) $\to_{}^{}$ AgCl(s)

Given $E °_{Cl 2 / {Cl}^{-}} = + 1.36 V, E °_{AgCl / Ag, {Cl}^{-}} = 0.22 V$

If $P_{{Cl}_{2}}$ = 1 atmand T = 298 K

A.

110 kJ/mol

B.

220 kJ/mol

C.

55kJ/mol

D.

1000 kJ/mol

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Explanation

AgCl(s) + e $\to_{}^{}$ Ag(s) + Cl^–E° = 22 V

1/₂Cl₂ + e $\to_{}^{}$ Cl^–E° = 1.36 V

We get

Ag(s) + $\frac{1}{2}$ Cl₂(g) $\to_{}^{}$ AgCl(s)E°_cell = 1.14 V

∴ ΔG = –nEF° = – (96500) (1.4) = –110 KJ/mol

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Which of these ions Cu⁺, Co³⁺, Fe²⁺ is stable in aqueous medium.

Given :E°_Cu2+/Cu+ = 0.15 volt ; E°_Cu+/Cu = 0.53 V ;E°_Co3+/Co2+ = 1.82 V ;

E°_Fe3+/Fe2+ = 0.77 V ;E°_Fe2+Fe = –0.44 V ;E°_O2,H+/H2O = 1.23 V

A.

Cu⁺, Co³⁺

B.

Co³⁺

C.

Cu⁺

D.

Co³⁺, Cu⁺, Fe²⁺

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Explanation

if reduction potential of metal ion is greater then O₂/H₂O couple, the ion is stable in water. So Co³⁺ is stable in water

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Select the correct statement if –

E°_Mg2+/Mg = –2.4V, E°_Sn4+/Sn2+ = 0.1 V, E°_{MnO4–,H+/Mn2+} = 1.5 V, E° _I2/I–= 0.5 V

Here,

A.

MnO₄^–is the strongest Oxidizing Agent and Mg is the strongest Reducing Agent.

B.

Sn⁴⁺ + 2I^– $\to_{}^{}$ Sn²⁺ + I₂ is a spontaneous reaction.

C.

Mg²⁺ + Sn²⁺ $\to_{}^{}$ Mg + Sn⁴⁺ is a spontaneous reaction.

D.

Weakest oxidizing agent is Sn⁴⁺ and weakest reducing agent is Mn²⁺.

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Explanation

[Hint: Reverse of (2) & (3) is spontaneous; weakest Oxidizing Agent here is Mg²⁺]

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The temperature coefficient of a standard Cd–cell is –5.0 $\times$ 10^–5 Vk^–1 whose emf at 25°C is 1.018 V. During the cell operation, the temperature will –

A.

increase

B.

decrease

C.

either

D.

remains constant

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Explanation

âˆµ ΔH = ΔG + T ΔS

= –nFE_cell + nFT ${(\frac{\partial E}{\partial T})}_{p}$

= –199.35 KJ/mol [n = 2 for Cd²⁺ +2e^–→ Cd & putting other values]

NowΔH < O ⇒ Exothermic reaction

⇒ Heat will be released increasing the temperature.

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A cell Ag | Ag⁺ || Cu⁺⁺ | Cu initially contains 2M Ag⁺ and 2M Cu⁺⁺ ions. The charger in cell potential after the passage of 10 amp current for 4825 sec is:

A.

–0.000738 V

B.

–1.00738

C.

–0.0038 V

D.

none

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Explanation

Q = 10 × 4825 = 4825 = 48250 C

no. of mole = $\frac{48250}{96500}$ = 0.5

Ag + $\frac{1}{2}$ Cu⁺⁺ $\to_{}^{}$ Ag⁺ + $\frac{1}{2}$ Cu

2.00 2.00

2 – 0.25 2 + 0.50

E_cell –E°_cell – $\frac{0 .0591}{1}$ log $\frac{[{Ag}^{+}]}{[{Cu}^{+ +}]^{1 / 2}}$

E₁ = E°_ell – $\frac{0 .0591}{1}$ log $\frac{2 .00}{{(2 .00)}^{1 / 2}}$

E₂ = E°_ell – $\frac{0 .0591}{1}$ log $\frac{2 .50}{{(1 .75)}^{1 / 2}}$

ΔE = E₂ –E₁ = $\frac{0 .0591}{1}$ $[\log \sqrt{2} - \log \frac{2 .50}{\sqrt{1 .75}}] = \frac{0 .0591}{1}$ [log 1.41 – log 1.88]

= $\frac{0 .0591}{1}$ [0.1492 –0.2742] = – $\frac{0 .0591}{1}$ × 0.125 = –.00738 V

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For the cell (at 298 K)

Ag(s) | AgCl(s) | Cl^–(aq) || AgNO₃(aq) | Ag(s)

Which of the following is correct –

A.

The cell emf will be zero when [Ag⁺]_a = [Ag⁺]_c ([Ag⁺] in anodic compartment = [Ag⁺] in cathodic compartment)

B.

The amount of AgCl(s) precipitate in anodic compartment will decrease with the working of the cell.

C.

The concentration of [Ag⁺] = constant, in anodic compartment during working of cell.

D.

E_cell = E°_{Ag+ | Ag} –E°_{Cl–|AgCl|Ag} $− \frac{0 .059}{1} \log \frac{1}{{[{Cl}^{-}]}_{a}}$

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Explanation

Conductivity is high due to [H⁺]

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During an electrolysis of conc. H₂SO₄, perdisulphuric acid (H₂S₂O₈) and O₂ from in equimolar amount. The amount of H₂ that will form simultaneously will be (2H₂SO4 → H₂S₂O₈ + 2H⁺ + 2e^–)

A.

thrice that of O₂ in moles

B.

twice that of O₂ in moles

C.

equal to that of O₂ in moles

D.

half of that of O₂ in moles

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Explanation

Anode $\{\begin{matrix} 2 H_{2} {SO}_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-} \\ 2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-} \end{matrix}\}$

Cathode {2H₂O → H₂ + 2OH^––2e^–} × 3.

Net: 2H₂SO₄ + 8H₂O → H₂S₂O8 + O₂ + 3H₂ + 6H⁺ + 6OH^–

Hence ratio of $^{n} O_{2}$ and $^{n} H_{2}$ is 1: 3.

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In producing chlorine by electrolysis 100 kW power at 125V is being consumed.How much chlorine per minute is liberated (ECE of chlorine is 0.367 $\times 10^{- 6} k g C^{- 1}$ )

A.

$1.76 \times 10^{- 3} k g$

B.

$9.67 \times 10^{- 3} k g$

C.

$17.61 \times 10^{- 3} k g$

D.

$3.67 \times 10^{- 3} k g$

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Explanation

Mass of the substance deposited at the cathode

$m = z i t$

= $z (\frac{W}{V}) t$

=0.367 $\times 10^{- 6} \times \frac{100 \times 10^{3}}{125} \times 60$

= $17.6 \times 10^{- 3} k g$

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What will be the quantity of iron deposited by ferrous and ferric ion by 1F (Fe = 56)

A.

14g, 9.3 g

B.

28g, 18.6 g

C.

56g, 37.2g

D.

None of these

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Explanation

B). 2F = 1 mole Fe deposited from Fe²⁺

1 F = $\frac{1}{2}$ mol Fe deposited from F $e^{2 +} = 28 g$

3 F = 1 mol Fe deposited from F $e^{3 +}$

1 F = $\frac{1}{3}$ mol Fe deposited from F $e^{3 +} = 18.6 g$

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