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The solubility product of silver iodide is $8.3 \times 10^{- 17}$ and the standard potential (reduction) of Ag, $A g^{+}$ electrode is + 0.800 volts at $25^{°} C .$ The standard potential of Ag, AgI/ $I^{-}$ electrode (reduction) from these data is-

– 0.30 V

+ 0.15 V

+ 0.10 V

– 0.15 V

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Explanation

(D). Solubility product reaction is

$A g I \to A g^{+} + I^{-}$

By calculating the EMF of this cell reaction from the given data and relating to $K_{s p}$ via the $∆ G °$ of the reaction, we can obtain $K_{s p}$ .

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How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere-

26.805 h

53.61 h

107.22 h

None of these

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Explanation

(B) $2 O H^{-} \to H_{2} O + \frac{1}{2} O_{2} + 2 e^{-} (A n o d e) 2 F = 1 m o l H_{2} O d e c o m p o s e d; s o f o r 2 m o l o f H_{2} O, 4 F o f e l e c t r i c i t y r e q u i r e d . Q = i \times t (i = 2 a m p .) t = \frac{4 \times 96500}{2} = 193000 s e c = 53.61 h o u r s .$

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A certain current liberates 0.504 g of hydrogen in 2 hours. How many gram of copper can be liberated by the same current flowing for the same time in aqueous $C u S O_{4}$ solution :

12.7

31.8

63.5

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Explanation

(B) $E q . o f H_{2} = E q . o f C u ∴ \frac{0.154}{1} = \frac{w}{63.5 / 2} ∴ w_{C u} = 16 g$

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Calculate the maximum work that can be obtained from the Daniell cell given below -

$Z n (s) | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u (s) .$

Given that $E_{Z n^{2 +} / Z n}^{°} = - 0.76 V a n d E_{C u^{2 +} / C u}^{°} = + 0.34 V$

– 212300 J

– 202100 J

– 513100 J

– 232120 J

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Explanation

(A) Cell reaction is : $Z n (s) + C u^{2 +} (a q) \to C u (s) + Z n^{2 +} (a q)$

Here n=2

${E^{\circ}}_{c e l l} = {E^{\circ}}_{c a t h o d e} - {E^{\circ}}_{a n o d e}$

(On the basis of reduction potential)

= + 0.34 – (– 0.76) = 1.10 V

We know that : $W_{m a x} = ∆ G^{\circ} = - n F E^{\circ}$

$= - (2 m o l) \times (96500 C m o l) \times (1.10 V) = - 212300 J$

$o r W_{m a x} = - 212300 J .$

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We have taken a saturated solution of $A g B r, K_{s p}$ of AgBr is $12 \times 10^{- 14} .$ If $10^{- 7}$ mole of $A g N O_{3}$ are added to 1 litre of this solution then the conductivity of this solution in terms of $10^{- 7}$ $S m^{- 1}$ units will be

$[λ °_{(A g^{+})} = 4 \times 10^{- 3} S m^{2} m o l^{- 1}; λ °_{(B r^{-})} = 6 \times 10^{- 3} S m^{2} m o l^{- 1}, λ °_{(N O_{3}^{-})} = 5 \times 10^{- 3} S m^{2} m o l^{- 1}]$

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Explanation

(A). The solubility of AgBr in presence of $10^{- 7}$ molar $A g N O_{3} i s 3 \times 10^{- 7} M .$

Therefore $[B r^{-}] = 3 \times 10^{- 4} m^{3}, [A g^{+}] = 4 \times 10^{- 4} m^{3} a n d [N {O_{3}}^{-}] = 10^{- 4} m^{3}$

Therefore $k_{t o t a l} = k_{B r^{-}} + k_{A g^{+}} + k_{N O_{3}^{-}} = 55 S m^{- 1}$

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The standard reduction potentials, $E^{\circ}$ , for the half reactions are as $Z n = Z n^{2 +} + 2 e; E ° = + 0.76 V$ and $F e = F e^{2 +} + 2 e; E ° = + 0.41$ V; the e.m.f. for the cell reaction, $F e^{2 +} + Z n = Z n^{2 +} + F e i s -$

– 0.35 V

+ 0.35 V

+ 1.17 V

– 1.17 V

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Explanation

(B). Since oxidation potential of Zn is higher than Fe so it will act as anode simultaneously Fe will act as cathode so EMF of cell will be Eº (ox.) of Zn + Eº (red) of Fe.

E = 0.76 + ( - 0.41) = + 0.35 V

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The calomel electrode is reversible with respect to-

Mercury

$H^{+}$

$H g^{2 +}$

$C l^{-}$

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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)

5.6 amp

7.2 amp

8.85 amp

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 :

0.18V

0.52V

0.82V

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:

96,500 C

9,650 C

965 C

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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