The amount of Ag (atomic mass = 108) deposite | vedclass

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(A) The reaction at the cathode is: $Ag^+ + e^- \rightarrow Ag(s)$.According to Faraday's law of electrolysis,the mass $W$ deposited is given by: $W =

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The reaction at the cathode is: $Ag^+ + e^- ightarrow Ag(s)$.According to Faraday's law of electrolysis,the mass $W$ deposited is given by: $W = \frac{M 	imes I 	imes t}{n 	imes F}$.Given: $M = 108 \, g/mol$,$I = 0.5 \, A$,$t = 1 \, h = 3600 \, s$,$n = 1$ (for $Ag^+$),and $F \approx 96500 \, C/mol$.Substituting the values: $W = \frac{108 	imes 0.5 	imes 3600}{1 	imes 96500}$.$W = \frac{194400}{96500} \approx 2.014 \, g$.Thus,the amount is closest to $2 \, g$.

The amount of $Ag$ (atomic mass $= 108$) deposited at the cathode when a current of $0.5 \, A$ is passed through a solution of $AgNO_3$ for $1 \, h$ is closest to $... \, g$.

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