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Chemistry NEET Questions — Free MCQ Practice

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The calomel electrode is reversible with respect to-

A.

Mercury

B.

$H^{+}$

C.

$H g^{2 +}$

D.

$C l^{-}$

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Explanation

(D). Since half cell reaction;

$2 e + H g_{2} C l_{2} (a q .) ⇌ 2 H g (l) + 2 C l^{-}$

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Zn amalgam is prepared by electrolysis of aqueous $Z n C l_{2}$ using Hg cathode (9 gm). How much current is to be passed through $Z n C l_{2}$ solution for 1000 seconds to prepare a Zn Amalgam with 25% Zn by wt. (Zn = 65.4)

A.

5.6 amp

B.

7.2 amp

C.

8.85 amp

D.

11.2 amp

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Explanation

C). Let x gm of Zn deposit on 9 gm of Hg

% of Zn in Amalgam $= \frac{x}{9 + x} \times 100 = 25 ∴ x = 3 g m$

Eq. of Zn= $\frac{3 \times 2}{65.4}; C u r r e n t = \frac{6}{65.4} \times \frac{96500}{1000} = 8.85 a m p .$

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The standard oxidation potentials of $C u / C u^{2 +}$ and $C u^{+} / C u^{2 +} a r e - 0.34 V a n d - 0.16 V$ respectively. The standard electrode potential of $C u^{+} / C u$ would be :

A.

0.18V

B.

0.52V

C.

0.82V

D.

0.49V

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Explanation

(B). Reactions $C u^{2 +} + 2 e^{-} \to C u; ∆ G ° = - n F E °$

$C u^{+} \to C u^{2 +} + e^{-}; ∆ G °_{I} = - 2 F \times 0.34 = 0.68 F$

$∆ G °_{I I} = F \times 0.16 = 0.16 F$

Adding, we get

$C u^{+} + e^{-} \to C u$

$∆ G °_{I I I} = ∆ G °_{I} + ∆ G °_{I I} = 0.52 F = - F E ° ∴ E ° = 0.52 V$

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Acidified water is electrolysed using an inert electrode. The volume of gases liberated at STP is 0.168L. The quantity of charge passed through the acidified water would be:

A.

96,500 C

B.

9,650 C

C.

965 C

D.

168 C

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Explanation

(C) $\underset{2 x}{2 H_{2} O} \to \underset{x}{2 H_{2 (g)}} + O_{2 (g)} ∴ 3 x = 0.168 ∴ x = 0.056 L V_{H_{2}} = 2 x = 0.112 L, V_{O_{2}} = x = 0.056 L 11.2 L o f H_{2} a t S T P \equiv 1 F 0.112 L o f H_{2} a t S T P \equiv 0.01 F 0.056 L o f O_{2} a t S T P = 0.01 F$

$∴$ The amount of electricity passed = 0.01F = 965C

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$E^{0} f o r t h e r e a c t i o n F e + Z n^{2 +} = Z n + F e^{2 +} i s - 0.35 V . T h e g i v e n c e l l r e a c t i o n i s -$

A.

Feasible

B.

Not feasible

C.

In equilibrium

D.

None

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Explanation

(B). Since EMF of the cell is negative i.e. Free energy change will be positive so cell reaction will not be feasible.

● ● ● ● ●

How much charge should be supplied to a cell for the electrolytic production of 245 gm $N a C l O_{4}$ from $N a C l O_{3}$ if the anode efficiency for the required reaction is 60%?

A.

6.43 × $10^{5} C$

B.

6.67 F

C.

6.43 × $10^{6} C$

D.

66.67 F

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Explanation

(A) $C l {O_{4}}^{-} + 2 H^{+} + 2 e^{-} \to C l {O_{3}}^{-} + H_{2} O$

Number of equivalents of $N a C l O_{4} = \frac{245}{61.25} = 4 \equiv 4 F$

No. of Faradays $= \frac{4 \times 100}{60} = 6.67 F = 6.43 \times 10^{5} C$

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During an electrolysis of conc. $H_{2} S O_{4}$ , perdisulphuric acid $(H_{2} S_{2} O_{8}) a n d O_{2}$ form in equimolar amount. The amount of $H_{2}$ that will form simultaneously will be

$(2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-})$

A.

thrice that of $O_{2}$ in moles

B.

twice that of $O_{2}$ in moles

C.

equal to that of $O_{2}$ in moles

D.

half of that of $O_{2}$ in moles

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Explanation

(A) Anode $\{\begin{matrix} 2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-} \\ 2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-} \end{matrix}\}$

Cathode $\{2 H_{2} O \to H_{2} + 2 O H - 2 e^{-}\} \times 3__________________________________$

$N e t : 2 H_{2} S O_{4} + 8 H_{2} O \to H_{2} S_{2} O_{8} + O_{2} + 3 H_{2} + 6 H^{+} + 6 O H^{-}$

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

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The specific conductivity of solution depends upon :

A.

Number of ions as well as mobility of ions

B.

Number of ions per c.c solution

C.

Number of ions per c.c as well as mobilities of ions

D.

Mobilities of ions only

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Explanation

(C). Specific conductance is the conductance per c.c. solution

● ● ● ● ●

Value of ${\land_{m}}^{0}$ for $S r C l_{2}$ in water at $25 ° C$ from the following data

Conc. (mol/It) 0.25 1

$\land_{m} (Ω^{- 1} c m^{2} m o l^{- 1})$ 260 250

A.

270

B.

260

C.

250

D.

255

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Explanation

(A). $\land_{m} = {\land_{m}}^{0} - b \sqrt{C} 260 = {\land_{m}}^{0} - 0.5 b . . . . . (1) 250 = {\land_{m}}^{0} - b . . . . . . (2) O n s o l v i n g (1) & (2), w e g e t {\land_{m}}^{0} = 270 Ω^{- 1} c m^{2} m o l^{- 1}$

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Calculate the energy obtainable from a lead storage battery in which 0.1 mol lead is consumed. Assume a constant concentration of 10.0 M $H_{2} S O_{4}$

A.

42.8 kJ

B.

12.8 kJ

C.

32.3 kJ

D.

20.2 kJ

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Explanation

(A) $P b O_{2} + 4 H^{+} + S {O_{4}}^{2 -} + 2 e^{-} \to P b S O_{4} + 2 H_{2} O; E º = 1.70 P b + S {O_{4}}^{2 -} \to P b S O_{4} + 2 e^{-}; E º = 0.3 P b O_{2} + P b + 4 H^{+} + 2 S {O_{4}}^{2 -} \to 2 P b S O_{4} + 2 H_{2} O; E º = 2.01$

$E = E ° - (\frac{0.0592}{2}) (\log \frac{1}{{[H^{+}]}^{4} {[S O_{4}^{2 -}]}^{2}}) = 2.01 - (\frac{0.0592}{2}) (\log \frac{1}{{(20)}^{4} {(10)}^{2}}) = 2.22 V$

Now, (0.100 mol Pb) $(\frac{2 m o l e^{-}}{m o l P b}) (\frac{95500 C}{m o l e^{-}}) = 19300 C$

Energy = qE = (19300 C) (2.22 V) = 42.8 kJ.

Chemistry MCQs for NEET — Practice Questions with Answers

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