When two electrolytic cells containing NiSO_4 | vedclass

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(A) Moles of $Cu = \frac{	ext{Mass of } Cu}{	ext{Atomic mass of } Cu} = \frac{1.6}{63.5} = 0.0252 \ mol$. Since the cells are connected in series,th

Vedclass · Accepted Answer

$1.48 \ g \ Ni, \ 5.44 \ g \ Ag$

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(A) Moles of $Cu = \frac{	ext{Mass of } Cu}{	ext{Atomic mass of } Cu} = \frac{1.6}{63.5} = 0.0252 \ mol$. Since the cells are connected in series,the same amount of charge $(Q)$ passes through each cell. According to Faraday's laws: $Cu^{2+} + 2e^- ightarrow Cu$ ($2 \ mol \ e^-$ produces $1 \ mol \ Cu$) $Ni^{2+} + 2e^- ightarrow Ni$ ($2 \ mol \ e^-$ produces $1 \ mol \ Ni$) $Ag^+ + e^- ightarrow Ag$ ($1 \ mol \ e^-$ produces $1 \ mol \ Ag$) For the same charge,the moles of metal produced are proportional to their equivalent weights. Moles of $Ni = 	ext{Moles of } Cu = 0.0252 \ mol$. Mass of $Ni = 0.0252 \ mol 	imes 58.7 \ g/mol = 1.48 \ g$. Moles of $Ag = 2 	imes 	ext{Moles of } Cu = 2 	imes 0.0252 = 0.0504 \ mol$. Mass of $Ag = 0.0504 \ mol 	imes 108 \ g/mol = 5.44 \ g$.

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