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Electrochemistry NEET Questions — Free MCQ Practice

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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.

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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

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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.

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Adiponitrile is manufactured electrolytically from acrylonitrile $C H_{2} = C H C N \to C N - (C H_{2})_{4} - C H$

How many kg of adiponitrile (molecular mass = 108) is produced in 9.65 hr using a current of 3750 A with 80% efficiency ?

A.

30 kg

B.

58 kg

C.

60 kg

D.

80 kg

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Explanation

(B) $2 C H_{2} = C H C N + 2 H^{+} + 2 e^{-} \to C N - (C H_{2})_{4} C N$

$m = \frac{(\frac{108}{2}) \times 10^{- 3} \times 3000 \times 9.65 \times 3600}{96500} = 58.32 k g$

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A certain metal salt solutions is electrolysed in series with a silver coulometer. The weights of silver and the metal deposited are 0.5094 g and 0.2653 g. Calculate the valency of the metal if its atomic weight is nearly that of silver.

A.

1

B.

2

C.

3

D.

4

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Explanation

(B) Evidently $\frac{0.5094}{0.2653} = \frac{E q u i v a l e n t w e i g h t o f A g}{E q u i v a l e n t w e i g h t o f m e t a l} \approx 0.2$

$∴ v a l e n c y r a t i o = \frac{V a l e n c y o f m e t a l}{V a l e n c y o f A g} = 2$

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The reduction potential of a half-cell consisting of a Pt electrode immersed in $1.5 M F e^{2 +}$ and $0.015 M F e^{3 +}$ solution at $25 ° C i s (E_{(A g^{+}, A g)}^{°} = + 0.770 v)$ .

A.

0.652 V

B.

0.88 V

C.

0.710 V

D.

0.850 V

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Explanation

(A) $E = 0.77 - \frac{0.059}{1} \log \frac{[F e^{2 +}]}{[F e^{3 +}]} = 0.77 - 0.059 \log \frac{1.5}{0.015} = 0.77 - 0.118 = 0.652 V$

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In a concentration cell, $Z n / Z n^{2 +} (1.0 M) | | Z n^{2 +} (0.15 M) / Z n$ as the cell discharges,

A.

reaction proceeds to the right

B.

the two solutions approach each other in concentration

C.

no reaction takes place

D.

water gets decomposed

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Explanation

(B). In the above representation of the cell, the reaction proceeds to the left; 1.0 M solution is used up.

Electrochemistry MCQs for NEET — Chemistry Questions with Answers

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