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Secondary Cells

Label the giv...

Question

Label the given diagram showing lead storage battery.

P b

, q-

P b O_{2}

, r-

5

H_{2} S O_{4}

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P b O_{2}

, q-

P b

, r- conc.

H_{2} S O_{4}

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P b_{3} O_{4}

, q-

P b O_{2}

, r-

50 %

H_{2} S O_{4}

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P b O_{2}

, q-

P b

, r- dil.

38 %

H_{2} S O_{4}

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Solution

The correct option is D p- $P b O_{2}$ , q- $P b$ , r- dil. $38 %$ $H_{2} S O_{4}$
Anode: $P b + S O_{4}^{2 -} ⟶ P b S O_{4} + 2 e^{-}$

Cathode: $P b O_{2} + 4 H^{+} + S O_{4}^{2 -} + 2 O^{-} ⟶ P b S O_{4} + 2 H_{2} O$

Therefore, At cathode, $p$ is $P b O_{2}$ .

At anode, $q$ is $P b$ .

$r$ is 38% $H_{2} S O_{4}$ (Sulphuric acid).
Hence option (D) is correct.

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Similar questions

During discharging of lead-storage acid battery following reaction takes place:

P b (s) + P b O_{2} (s) + 2 H_{2} S O_{4} \to 2 P b S O_{4} (s) + 2 H_{2} O

If 2.5 amp of current is drawn for 965 minutes, $H_{2} S O_{4}$ consumed is:

H_{2} S O_{4}

1.294

1.139

. S u l p h u r i c a c i d o f d e n s i t y

i s

b y w e i g h t a n d t h a t o f d e n s i t y

i s

b y w e i g h t . T h e b a t t e r y h o l d

o f t h e a c i d a n d v o l u m e r e m a i n s p r a c t i c a l l y c o n s t a n t d u r i n g d i s c h a r g e . C a l c u l a t e a m p e r e - h o u r o f w h i c h t h e b a t t e r y m u s t h a v e b e e n u s e d . T h e c h a r g i n g a n d d i s c h a r g i n g r e a c t i o n s a r e :

(c h a r g i n g)

The contrapositive of $(\sim p \land q) \to (q \land \sim r)$ is

[P] B r_{2} ⟶ C_{2} H_{4} B r_{2} N a N H_{2} ⟶ N H_{3} Q

[Q] 20 % H_{2} S O_{4} ⟶ H g^{2 +}, Δ [R] Z n - H g / H C l ⟶ [S]

P b (s) + P b O_{2} + 2 H_{2} S O_{4} (a q) \to 2 P b S O_{4} (s) + 2 H_{2} O

1.25 A

H_{2} S O_{4}

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