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Chemistry NEET Questions — Free MCQ Practice

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

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

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A flashlight cell has the cathodic reaction $2 M n O_{2} (s) + Z n^{+ 2} + 2' e^{-} \to Z n M n_{2} O_{4} (S)$

If the flashlight cell is to give out 4.825 mA, how long could it run if initially 8.7 g of the limiting reagent $M n O_{2}$ is present ? [Mn = 55, O = 16]

A.

$2 \times 10^{6} s e c$

B.

$4 \times 10^{6} s e c$

C.

$6 \times 10^{6} s e c$

D.

$8 \times 10^{6} s e c$

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Explanation

(A) $\frac{8.7}{87} \times 96500 = 4.825 \times 10^{- 3} 10^{- 3} \times t \Rightarrow t = 2 \times 10^{6} s e c$

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Calculate the cell EMF in mV for $P t | H_{2} (1 a t m) | H C l (0.01 M) | A g C l (s) | A g (s) a t 298 K$

If $∆ {G_{r}}^{°}$ values are at $25 ° C$ $- 109.56 \frac{k J}{m o l} f o r A g C l (s) & - 130.79 \frac{k J}{m o l} f o r (H^{+} + C l^{-}) (a q)$

A.

456 mV

B.

654 mV

C.

546 mV

D.

None of these

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Explanation

(A)

$∆ G_{c e l l r e a c t i o n}^{°} = 2 (- 130.79) - 2 (- 109.56) = 42.46 k J / m o l e (f o r H_{2} + 2 A g C l \to 2 A g + 2 H^{+} + 2 C l^{-}) ∴ E °_{c e l l} = \frac{- 42460}{- 2 \times 96500} = + 0.220 V N o w, E_{c e l l} = + 0.220 + \frac{0.059}{2} \log \frac{1}{{(0.01)}^{4}} = 0.456 V = 456 m V$

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A current of 0.1A was passed for 2hr through a solution of cuprocyanide and 0.3745 g of copper was deposited on the cathode. Calculate the current efficiency for the copper deposition. (Cu – 63.5)

A.

79%

B.

39.5%

C.

63.25%

D.

63.5%

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Explanation

(A) $m (t h e o r e t i c a l) = \frac{63.5 \times 0.1 \times 7200}{96500} = 0.4738 g ∴ % e f f i c i e n c y = \frac{0.3745}{0.4738} \times 100 = 79 %$

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In acidic medium $M n {O_{4}}^{-}$ is an oxidising agent. $M n {O_{4}}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O . I f H^{+}$ ion concentration is doubled, electrode potential of the half cell will :

A.

Increase by 28.46 mV

B.

Decrease by 28.46 mV

C.

Increase by 14.23 mV

D.

Decrease by 142.30 mV

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Explanation

(A) $E = E^{°} + \frac{0.0591}{5} \log {(2)}^{8} o r E - E^{°} = \frac{0.0591}{5} \times 8 \log 2 = 0.02846 V = 28.46 m V$

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Consider the cell $A g (s) | A g B r (s) | B r^{-} (a q) | | A g C l (s) | C l^{-} (a q) | A g (s) a t 25 ° C .$ The solubility product constants of AgBr & AgCl are respectively $5 \times 10^{- 13} & 1 \times 10^{- 10} .$ For what ratio of the concentrations of $B r^{-} & C l^{-}$ ions would the emf of the cell be zero ?

A.

1 : 200

B.

1 : 100

C.

1 : 500

D.

200 : 1

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Explanation

(A) $E_{B r^{-} / A g B r / A g}^{0} = E_{A g^{+} / A g}^{0} + \frac{0.059}{1} \log K_{S P} A g B r = E_{A g^{+} / A g}^{0} = - 0.7257 a n d E_{C l^{-} / A g C l / A g}^{0} = E_{A g^{+} / A g}^{0} + \frac{0.059}{1} \log K_{S P} A g C l = E_{A g^{+} / A g}^{0} = - 0.59 N o w c e l l r e a c t i o n i s A g + B r^{-} \to A g B r + l e^{-} A g C l + l e^{-} \to A g + C l^{-} B r^{-} + A g C l \overset{l e^{-}}{\to} C l^{-} + A g B r 0 = (0.7257 - 0.59) + \frac{0.059}{1} \log \frac{[B r^{-}]}{[C l^{-}]} \Rightarrow \frac{[B r^{-}]}{[C l^{-}]} = 0.005$

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Calculate the EMF of the cell at 298 K $P t | H_{2} (1 a t m) | N a O H (x M), N a C l (x M) | A g C l (s) | A g$ ${E^{°}}_{C l^{-} / A g C l / A g} = + 0.222 V$

A.

– 0.604 V

B.

– 0.04 V

C.

1.048 V

D.

emf depends on x and cannot be determined unless value of x is given

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Explanation

(A) Anode : $½ H_{2} \to H^{+} + 1 e^{-}$

Aathode : $A g C l + 1 e^{-} \to A g + C l^{-}$

$N e t : \frac{1}{2} H_{2} + A g C l \overset{1 e^{-}}{\to} H^{+} + A g + C l^{-} E_{c e l l} = + 0.222 + \frac{0.059}{1} \log \frac{1}{[H^{+}] [C l^{-}]} = + 0.222 + 0.059 \log \frac{[O H^{+}]}{[10^{- 14}] [C l^{-}]} = + 0.222 + 0.059 (14) = + 1.048 v o l t$

Chemistry MCQs for NEET — Practice Questions with Answers

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