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After the electrolysis of conc. $H_{2} {SO}_{4}$ solution , perdisulphuric acid $(H_{2} S_{2} O_{8})$ and $O_{2}$ form in equimolar amount. than the amount of $H_{2}$ that will form simultaneously is- $(2 H_{2} {SO}_{4} ⟶ H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-})$

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Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing

0.2

Faraday electricity through an aqueous solution of potassium succinate, what is the total volume of gases (at both cathode and anode) at STP (

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and

273 K

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The density of metal and equivalent weight of a metal are

10.5 g {cm}^{- 3}

and

100

, respectively. The time required for a current of

3 amp

to deposit a

0.005 mm

thick layer of the same metal on an area of

80 {cm}^{2}

is closest to

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How long (approximate) should water be electrolysed by passing through

100

amperes

current so that the oxygen released can completely burn

27.66 g

of diborane? (Atomic weight of

B = 10.8 u

)

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Chemistry>Physical Chemistry>Electrochemistry>Electrolytic Cells and Electrolysis

When

0.1 m o l {M n O}_{4}^{2 -}

is oxidized the quantity of electricity required to completely

M n O_{4}^{2 -}

M n O_{4}^{-}

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The number of electrons delivered at the cathode, during electrolysis by a current of

1 ampere in 60 seconds

is (charge on electron

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Chemistry>Physical Chemistry>Electrochemistry>Electrolytic Cells and Electrolysis

The weight of silver (at.wt.

= 108

) displaced by a quantity of electricity which displaces

5600 m L

O_{2}

STP

will be

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A current of $10.0 A$ flows for $2.00 h$ through an electrolytic cell containing a molten salt of metal $X$ . This results in the decomposition of $0.250 mol$ of metal $X$ at the cathode. The oxidation state of $X$ in the molten salt is:

$(F = 96, 500 C)$

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Chemistry>Physical Chemistry>Electrochemistry>Electrolytic Cells and Electrolysis

The quantity of electricity needed to separately electrolyse

1 M

solution of

{ZnSO}_{4}, {AlCl}_{3}

and

{AgNO}_{3}

completely is in the ratio of

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When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is:

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For electroplating,

1.5 amp

current is passed for

250 s

through

250 m L of 0.15 M

solution of

M S O_{4} .

Only

85 %

of the current was utilised for electrolysis. The molarity of

M S O_{4}

solution after electrolysis is closest to [Assume that the volume of the solution remain constant]

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An aqueous solution of

{CuSO}_{4}

is subjected to electrolysis using inert electrodes. The

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of the solution will

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A constant current (0.5 amp) is passed for 1 hour through (i) aqueous

A g N O_{3}

. (ii) aqueous

C u S O_{4}

and (iii) molten

A l F_{3}

, separately. The ratio of the mass of the metals deposited on the cathode is

[M_{Ag}, M_{Cu}, M_{Al}

are molar masses of the respective metals

]

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One litre solution of

{MgCl}_{2}

is electrolyzed completely by passing a current of

1 A

for

16 \min 5

sec. The original concentration of

{MgCl}_{2}

solution was (Atomic mass of

Mg = 24

)

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A solution of

N i {(N O_{3})}_{2}

is electrolyzed between platinum electrode

0.1

Faraday electricity. How many mole of

N i

will be deposited at the cathode?

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The number of Faradays(F) required to produce

20 g

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) is

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What time in sec is required for depositing all the silver present in

125 ml

1 M {AgNO}_{3}

solution by passing a current of

241.25

A (Given

1 F = 96500 C

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0.5 F

of electricity is passed through

500 mL

of copper sulphate solution. The amount of copper which can be deposited will be

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Calculate the molarity of a solution containing

5 g

N a O H

dissolved in the product of a

H_{2} - O_{2}

fuel cell operated at

1 A

current for

595.1

hours.

(Assume

1 F = 96500 C / m o l

of electrons and molecular weight of

N a O H

40 g m o l^{- 1}

)

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During the electrolysis of molten sodium chloride, the time required to produce

0.10 mol

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Two Faraday of electricity is passed through a solution of

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. The mass of copper deposited at the cathode is: (Atomic mass of Cu = 63.5 amu)

After the electrolysis of conc. H2SO4 solution , perdisulphuric acid H2S2O8 and O2 form in equimolar amount. than the amount of H2 that will form simultaneously is- 2H2SO4⟶H2S2O8+2H++2e-

Important Questions on Electrochemistry

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After the electrolysis of conc. $H_{2} {SO}_{4}$ solution , perdisulphuric acid $(H_{2} S_{2} O_{8})$ and $O_{2}$ form in equimolar amount. than the amount of $H_{2}$ that will form simultaneously is- $(2 H_{2} {SO}_{4} ⟶ H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-})$