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Electrochemistry NEET Questions — Free MCQ Practice

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In a half-cell containing $[T l^{3 +}] = 0.1 M a n d [T l^{+}] = 0.01 M,$ the cell potential is – 1.2496 V for the reaction $T l^{+} \to T l^{3 +} + 2 e^{-}$ The standard reduction potential of the $T l^{+} / T l^{3 +}$ couple at $25 ° C$ is

A.

1.44 V

B.

0.61 V

C.

2.44 V

D.

1.22 V

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Explanation

(D) For the oxidation reaction,

$T l^{+} \to T l^{3 +} + 2 e^{-} E = E^{°} - \frac{R T}{n F} \log \frac{[o x i]}{[r e d]} - 1.2496 = E ° - \frac{0.059}{2} \log \frac{0.01}{0.01} ∴ E ° = 1.220 V$

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The standard reduction potentials for $M n^{3 +} / M n^{2 +}$ and $M n O_{2} / M n^{3 +}$ couples in acid medium at $25 ° C$ are 1.50 V and 1.00 V respectively. The following reaction:

$2 M n^{3 +} + 2 H_{2} O \to M n^{2 +} + M n O_{2} + 4 H^{+} i s$

A.

reversible

B.

non-spontaneous

C.

spontaneous

D.

irreversible

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Explanation

(C) $M n^{3 +} + e^{-} \to M n^{2 +}; E ° = 1.50 V M n^{3 +} + 2 H_{2} O \to M n O_{2} + 4 H^{+} + e^{-}; E ° = - 1.0 V$

Eº for the above reaction=1.50 V + (– 1.0)V = 0.50 V

Since Eº is positive, $∆$ Gº is negative and the reaction is spontaneous.

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If a 100 mL solution of 0.1M HBr is titrated using a very concentrated solution of NaOH, then the conductivity (specific conductance) of the solution at the equivalence point will be (assume volume change is negligible due to addition of NaOH). Calculate your answer in terms of $10^{- 1} S m^{- 1},$

$[G i v e n {\land^{°}}_{(N a^{+})} = 8 \times 10^{- 3} S m^{2} m o l^{- 1} {\land^{°}}_{(B r^{-})} = 4 \times 10^{- 3} S m^{2} m o l^{- 1}]$

A.

6

B.

12

C.

15

D.

24

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Explanation

(B) At the equivalence point the concentrations will be $[B r^{-}] = 100 m^{3}, [N a^{+}] = 100 m^{3}$

Therefore,

$K_{t o t a l} = k_{B r^{-}} + k_{N a^{+}} = 1.2 S m^{- 1} = 12 \times 10^{- 1} S m^{- 1}$

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The volume of gases liberated at STP when a charge of 2F is passed through aqueous solution of sodium phosphate, is :

A.

11.2L

B.

44.8L

C.

33.6L

D.

22.4L

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Explanation

(C) At anode $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$

At cathode $2 H_{2} O + 2 e^{-} \to 2 O H^{-} + H_{2}$

After passage of 2F, one mole of $H_{2}$ and 1/2 mole of $O_{2}$ would be produced. The total volume is 33.6L.

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It is desired to convert the energy derived from the combustion of propane into electrical energy, via a fuel cell. Given that the standard free energies of formation of $C_{3} H_{8} (g), H_{2} O (l) a n d C O_{2} (g)$ are -23.5, -237.2 and -394.4 $k J m o l^{- 1}$ respectively, calculate the standard EMF of the propane fuel cell.

A.

1.10 V

B.

2.10V

C.

3.12 V

D.

4.12 V

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Explanation

(A) For the combustion of propane

$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} (g) + 4 H_{2} O (l)$ the overall $∆ G °$ for the reaction

$= 3 x + 394.4 + 4 x - 237.2 = - [- 23.5 + 0] = - 1183.2 - 948.8 + 23.5 = - 2108.5 k J C_{3} H_{8} (g) + 6 H_{2} O (l) \to 3 C O_{2} (g) + 20 H = 20 e \frac{5 O_{2} (g) + 2 O H^{+} + 20 e \to 10 H 2 O (l)}{C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)}$

$∴$ 20 electrons are involved in the cell reactions

$∆ G^{0} = - n F E^{0}; E^{0} = - \frac{∆ G^{0}}{n F} = \frac{+ 21085 \times 10^{3}}{20 \times 96500} = 1.10 V$

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Na-amalgam is prepared by electrolysis of NaCl solution using liquid Hg as cathode. How long should the current of 10 amp. is passed to produce 10% Na – Hg on a cathode of 10 gm Hg. (atomic mass of Na = 23).

A.

7.77 min.

B.

9.44 min.

C.

5.24 min.

D.

11.39 min.

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Explanation

(A) 90 gm Hg has 10 gm Na

$∴ 10 g m H g = \frac{10}{90} \times 10 = \frac{10}{9} g m N a ∴ w e i g h t o f N a = \frac{M}{n} \times \frac{i \times t}{96500} \frac{10}{9} = \frac{23}{1} \times \frac{10 \times t}{96500} [∴ N a^{+} + e \to N a] = \frac{10 \times 96500}{9 \times 10 + 23} = 7.77 m i n$

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The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is 0.79 V. The maximum amount of AgCl that can dissolve in $10^{6} L o f a 0.1 M A g N O_{3}$ solution is

A.

0.5 mmol

B.

1.0 mmol

C.

2.0 mmol

D.

2.5 mmol

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Explanation

(B) $A g C l + 1 e^{-} \to A g + C l^{-} E ° = - 0.2 V A g \to A g^{+} + 1 e^{-} E ° = - 0.2 V A g C l \overset{1 e^{-}}{\to} A g^{+} C l^{-} E ° = - 0.2 V E ° = \frac{0.059}{n} \log K \Rightarrow - 0.59 = \frac{0.059}{1} \log K_{s p} \Rightarrow K_{s p} = 10^{- 10}$

Now solubility of AgCl in 0.1 M AgNO₃

$S (S + 0.1) = 10^{- 10} \Rightarrow S = 10^{- 9} m o l / L$

Hence 1 mole dissolves in $10^{9}$ L solution hence in $10^{6}$ L amount that dissolves in 1 mmol.

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The $k = 4.95 \times 10^{- 5} S c m^{- 1}$ for a 0.00099 M solution. Calculate the reciprocal of the degree of dissociation of acetic acid, if $\land_{m}^{0}$ for acetic acid is 400 S $c m^{2} m o l^{- 1}$

A.

7

B.

8

C.

9

D.

10

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Explanation

(B) $λ_{M} = \frac{100 \times 4.95 \times 10^{- 5}}{0.00099} = 50 S c m^{2} m o l^{- 1} α = \frac{50}{400} = 0.125$

1/.125 = 8

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A hydrogen electrode X was placed in a buffer solution of sodium acetate and acetic acid in the ratio a : b and another hydrogen electrode Y was placed in a buffer solution of sodium acetate and acetic acid in the ratio b : a. If reduction potential values for two cells are found to be $E_{1} a n d E_{2}$ respectively w.r.t standard hydrogen electrode, the $p K_{a}$ value of the acid can be given as

A.

$\frac{E_{1} - E_{2}}{0.118}$

B.

$\frac{E_{1} + E_{2}}{0.118}$

C.

$\frac{E_{1}}{E_{2}} \times 0.118$

D.

$\frac{E_{2} - E_{1}}{0.118}$

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Explanation

(B) $H^{+} + e \to (1 / 2) H_{2} (g) E_{1} = 0 - 0.0591 \log \frac{1}{{(H^{+})}_{1}} E_{1} = 0 + 0.591 \log {[H^{+}]}_{1} = - 0.591 p H_{1} E_{2} = - 0.591 p H_{2} p H_{1} = p k_{a} + \log \frac{S a l t}{A c i d}; p H_{1} = p k_{a} + \log \frac{a}{b} . . . . . (1) p H_{1} = p k_{a} + \log \frac{b}{a}; p H_{2} = p k_{a} - \log \frac{a}{b} . . . . . . . . . (2) A d d (1) & (2) p H_{1} + p H_{2} = 2 p k_{a} 2 p k_{a} = - \frac{E_{1}}{0.0591} - \frac{E_{2}}{0.0591}; p k_{a} = - [\frac{E_{1} + E_{2}}{0.118}]$

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At what $\frac{[B r^{-}]}{\sqrt{[C O_{3}^{2 -}]}}$ does the following cell have its reaction at equilibrium ?

$A g (s) | A g_{2} C O_{3} (s) | N a_{2} C O_{3} (a q) | | K B r (a q) | A g B r (s) | A g (s)$

$K_{S P} = 8 \times 10^{- 12} f o r A g_{2} C O_{3} a n d K_{S P} = 4 \times 10^{- 13} f o r A g B r$

A.

$\sqrt{1} \times 10^{- 7}$

B.

$\sqrt{2} \times 10^{- 7}$

C.

$\sqrt{3} \times 10^{- 7}$

D.

$\sqrt{4} \times 10^{- 7}$

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Explanation

(B) anode : Ag(s) $\to A g^{+} (a q) + 1 e^{-}$

cathode: $A g^{+} (a q) + 1 e^{-} \to A g$

$N e t : A g^{+} (A g B r) \overset{1 e^{-}}{\to} A {g^{+}}_{(A g_{2} C O_{3})} 0 = 0 + \frac{0.059}{1} \log \frac{(\frac{K_{S P} A g B r}{[B r^{-}]})}{\sqrt{\frac{K_{S p} A g_{2} C O_{3}}{[C O_{3}^{2 -}]}}} \Rightarrow \frac{K_{S P} A g B r}{[B r^{-}]} = \sqrt{\frac{K_{S p} A g_{2} C O_{3}}{[C O_{3}^{2 -}]}} \Rightarrow \frac{4 \times 10^{- 13}}{\sqrt{8 \times 10^{- 2}}} = \frac{[B r^{-}]}{\sqrt{[C O_{3}^{2 -}]}} \Rightarrow \frac{[B r^{-}]}{\sqrt{[C O_{3}^{2 -}]}} = \sqrt{2} \times 10^{- 7}$

Electrochemistry MCQs for NEET — Chemistry Questions with Answers

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