On passing a current through KCl solution,19. | vedclass

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(A) The reactions at the cathode are:$K^+ + e^- \rightarrow K$$Al^{3+} + 3e^- \rightarrow Al$According to Faraday's laws of electrolysis,the mass depo

Vedclass · Accepted Answer

$4.5$

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(A) The reactions at the cathode are:$K^+ + e^- ightarrow K$$Al^{3+} + 3e^- ightarrow Al$According to Faraday's laws of electrolysis,the mass deposited is proportional to the equivalent mass.Moles of $K$ deposited $= 19.5 \ g / 39 \ g/mol = 0.5 \ mol$.Since $1 \ mol$ of $K^+$ requires $1 \ mol$ of electrons,$0.5 \ mol$ of electrons are passed.For $Al^{3+}$,the reaction is $Al^{3+} + 3e^- ightarrow Al$.Thus,$3 \ mol$ of electrons deposit $1 \ mol$ of $Al$.So,$0.5 \ mol$ of electrons will deposit $0.5 / 3 = 1/6 \ mol$ of $Al$.Mass of $Al$ deposited $= (1/6) \ mol 	imes 27 \ g/mol = 4.5 \ g$.

On passing a current through $KCl$ solution,$19.5 \ g$ of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride,the amount of aluminium deposited is ............ $g$

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