JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

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100 \mathrm{~g} \mathrm{~Al}

is to be deposited on the cathode of the electrolytic cell containing

\mathrm{Al}_2 \mathrm{O}_3

in the solvent

\mathrm{Na}_3 \mathrm{AlF}_6

by a current of

125 \mathrm{~A}

. If at the cathode, reaction is

\mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}

then time required for the passage of current is

(\mathrm{Al}=27)

View answer and explanation

Correct option: A

143 min

Option A (143 min) is correct because it is the time required to deposit 100g of Al, calculated using Faraday's laws of electrolysis.

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25^{\circ} \mathrm{C}

0.295 \mathrm{~V}

{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}}

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

{AlCl}_3

STP

({F}=96500 {C})

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

\mathrm{pH}=3

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

\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