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(A) In the electrolysis of an aqueous $NiI_2$ solution using $Ni$ electrodes,the process involves the dissolution of $Ni$ at the anode and the deposit

Vedclass · Accepted Answer

Increase by $29.5 \ g$

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(A) In the electrolysis of an aqueous $NiI_2$ solution using $Ni$ electrodes,the process involves the dissolution of $Ni$ at the anode and the deposition of $Ni$ at the cathode.At the anode: $Ni(s) ightarrow Ni^{2+}(aq) + 2e^-$.At the cathode: $Ni^{2+}(aq) + 2e^- ightarrow Ni(s)$.According to Faraday's law of electrolysis,the mass deposited or dissolved is given by $m = Z 	imes Q$,where $Z$ is the electrochemical equivalent and $Q$ is the charge in Faradays.For $Ni^{2+} + 2e^- ightarrow Ni$,the equivalent mass of $Ni$ is $\frac{	ext{Atomic Mass}}{	ext{n-factor}} = \frac{59}{2} = 29.5 \ g$.Passing $1$ equivalent of charge $(1 \ F)$ results in the deposition of $1$ equivalent mass of $Ni$ at the cathode.Therefore,the mass of the cathode increases by $29.5 \ g$.

In the electrolysis of an aqueous $NiI_2$ solution using $Ni$ electrodes,what happens to the mass of the cathode upon passing $1$ equivalent of charge? (Given: atomic mass of $Ni = 59$)

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