JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

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Application +4 | -1

A 4.0 molar aqueous solution of

\mathrm{NaCl}

is prepared and

500 \mathrm{~mL}

of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes. If the cathode is a

\mathrm{Hg}

electrode, the maximum weight

(\mathrm{g})

of amalgam

(\mathrm{Na}-\mathrm{Hg})

formed from this solution is : (atomic mass:

\mathrm{Na}=23, \mathrm{Hg}=200

)

View answer and explanation

Correct option: A

446

Option A (446) is correct because it is the mass of the Na-Hg amalgam formed based on the moles of Na+ reduced at the cathode.

More Questions from ELECTRO CHEMISTRY

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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