Function of fluorspar in electrolytic reduction of alumina dissolved in fused cryolite (N{a₃}Al{Ft

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p-Block Elements - I

medium

The function of fluorspar in the electrolytic reduction of alumina dissolved in fused cryolite $(N{a_3}Al{F_6})$ is

As a catalyst

To lower the temperature of the melt and to make the fused mixture very conducting

To decrease the rate of oxidation of carbon at the anode

None of the above

(IIT-1993)

Solution

Fluorspar $( CaF_2 )$ is added in small quantity in the electrolytic reduction of alumina dissolved in fused cryolite $( Na_3A l F_6)$. Addition of cryolite and fluorspar increases the electrical conductivity of alumina and lowers the fusion temperature to around $1140 \,K$.

Standard 11

Chemistry

The purification of alumina is called

easy

(AIIMS-1999)

Give reasons:

$(i)$ Conc. $HNO _{3}$ can be transported in aluminium container.

$(ii)$ $A$ mixture of dilute $NaOH$ and aluminium pieces is used to open drain.

$(iii)$ Graphite is used as lubricant.

$(iv)$ Diamond is used as an abrasive.

$(v)$ Aluminium alloys are used to make aircraft body.

$(vi)$ Aluminium utensils should not be kept in water overnight.

$(vii)$ Aluminium wire is used to make transmission cables.

medium

Boric acid is polymeric due to

easy

In the commercial electrochemical process for aluminium extraction, the electrolyte used is

medium

(IIT-1999)

${H_3}B{O_3}\,\xrightarrow{{{T_1}}}X\,\xrightarrow{{{T_2}}}Y\,\xrightarrow{{\operatorname{Re} d\,\,\,hot}}{B_2}{O_3}$

if $T_1 < T_2$ then $X$ and $Y$ respectively are

medium

The function of fluorspar in the electrolytic reduction of alumina dissolved in fused cryolite $(N{a_3}Al{F_6})$ is

Solution

Similar Questions

The purification of alumina is called

Boric acid is polymeric due to

In the commercial electrochemical process for aluminium extraction, the electrolyte used is

${H_3}B{O_3}\,\xrightarrow{{{T_1}}}X\,\xrightarrow{{{T_2}}}Y\,\xrightarrow{{\operatorname{Re} d\,\,\,hot}}{B_2}{O_3}$ if $T_1 < T_2$ then $X$ and $Y$ respectively are

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${H_3}B{O_3}\,\xrightarrow{{{T_1}}}X\,\xrightarrow{{{T_2}}}Y\,\xrightarrow{{\operatorname{Re} d\,\,\,hot}}{B_2}{O_3}$

if $T_1 < T_2$ then $X$ and $Y$ respectively are