Passage of electricity through dil ${H}_2 {SO}_4$ for 1 hour liberates a total of $224\, \mathrm{ml}$ of ${H}_2$ at S.T.P. The magnitude of current would be :-

JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

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Passage of electricity through dil

{H}_2 {SO}_4

for 1 hour liberates a total of

224\, \mathrm{ml}

{H}_2

at S.T.P. The magnitude of current would be :-

View answer and explanation

Correct option: B

0.54 amperes

Option B (0.54 amperes) is correct because it is calculated by using Faraday's laws of electrolysis to relate the amount of hydrogen liberated to the current passed.

More Questions from ELECTRO CHEMISTRY

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25^{\circ} \mathrm{C}

, the standard emf of cell having reactions involving two electron change is found to be

0.295 \mathrm{~V}

. The equilibrium constant of the reaction is Which are used as secondary reference electrodes? For the following cell with hydrogen electrodes at two different pressures

{p_1}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {p_2}

\mathop {\mathrm{Pt}\left( {{\mathrm{H}_2}} \right)}\limits_{{p_1}} \left| {\mathop {{\mathrm{H}^ + }\left( {aq} \right)}\limits_{1\mathrm{ }M} } \right|\mathop {Pt\left( {{\mathrm{H}_2}} \right)}\limits_{{p_2}}

emf is given by A current of strength 2.5 A was passed through

\mathrm{CuS}{\mathrm{O}_4}

solution for 6 min 26 s. The amount of copper deposited is (At. Wt. of Cu = 63.5, 1F = 96500 C) A current strength of 9.65 amperes is passed through excess fused

{AlCl}_3

for 5 hours. How many litres of chlorine will be liberated at

STP

({F}=96500 {C})

The electrode potential (in millivolts) of

2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{Ag}_2 \mathrm{S}(\mathrm{s})+2 \mathrm{e}^{-}

in a solution buffered at

\mathrm{pH}=3

and which is also saturated with

0.1 \mathrm{~M}-\mathrm{H}_2 \mathrm{S},

is (Given: for

\mathrm{H}_2 \mathrm{S}, K_{\mathrm{a}_1}=10^{-8} ; K_{\mathrm{a}_2}=10^{-13}, K_{\mathrm{sp}}

\mathrm{Ag}_2 \mathrm{S}=4 \times 10^{-48},

E^{\circ}_{{\mathrm{Ag} | \mathrm{Ag}}^{+}}=0.80 \mathrm{~V},

\log 2=0.3,

2.303 R T / F=0.06

)