Passage of a current for 548 , s through a si | vedclass

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(A) The reduction reaction at the cathode is: $Ag^{+} + e^{-} \longrightarrow Ag$According to Faraday's law,$1 \, F$ $(96500 \, C)$ of charge deposits

Vedclass · Accepted Answer

$1.22$

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(A) The reduction reaction at the cathode is: $Ag^{+} + e^{-} \longrightarrow Ag$According to Faraday's law,$1 \, F$ $(96500 \, C)$ of charge deposits $108 \, g$ of silver.Using the formula $Q = I 	imes t$,where $Q = n 	imes F$:$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.746 \, g}{108 \, g/mol} \approx 0.006907 \, mol$Charge $Q = 0.006907 \, mol 	imes 96500 \, C/mol \approx 666.56 \, C$Since $I = \frac{Q}{t}$,we have $I = \frac{666.56 \, C}{548 \, s} \approx 1.216 \, A \approx 1.22 \, A$.

Passage of a current for $548 \, s$ through a silver coulometer results in the deposition of $0.746 \, g$ of silver. What is the current (in $A$)?

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