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Question

(D) For $AgNO_3$: $Ag^{+} + e^{-} 	o Ag$. $1 	ext{ Faraday}$ of electricity deposits $1 	ext{ mole}$ of $Ag$ atoms. For $CuSO_4$: $Cu^{2+} + 2e^{-}

Vedclass · Accepted Answer

$2x = y$

Vedclass · Answer

(D) For $AgNO_3$: $Ag^{+} + e^{-} 	o Ag$. $1 	ext{ Faraday}$ of electricity deposits $1 	ext{ mole}$ of $Ag$ atoms. For $CuSO_4$: $Cu^{2+} + 2e^{-} 	o Cu$. $2 	ext{ Faraday}$ of electricity deposits $1 	ext{ mole}$ of $Cu$ atoms. Let the quantity of electricity passed be $Q 	ext{ Faraday}$. Number of moles of $Ag$ atoms $(x)$ = $Q/1 = Q$. Number of moles of $Cu$ atoms $(y)$ = $Q/2$. Comparing $x$ and $y$: $x = Q$ and $y = Q/2$. Therefore,$x = 2y$ or $2y = x$. Wait,re-evaluating: $x = Q$ and $y = Q/2$,so $x = 2y$ is not an option. Let's check the ratio: $x/y = Q / (Q/2) = 2$. Thus,$x = 2y$. Looking at the options provided,if $x$ is the number of atoms,$x = 2y$ is the correct relationship.

When the same quantity of electricity is passed through aqueous solutions of $AgNO_3$ and $CuSO_4$,if the number of atoms of $Ag$ and $Cu$ deposited are $x$ and $y$ respectively,then determine the correct relationship.

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