question_answer

Tarnished silver contains \[A{{g}_{2}}S\]. Can this tarnish be removed by placing tarnished silver ware in an aluminium pan containing an inert electrolyte solution such as NaCI? Given that the standard reduction potentials for the half reactions are:                                       \[A{{g}_{2}}S(s)\,+2{{e}^{-}}\xrightarrow{\,}\,2Ag(s);\]\[+{{S}^{2-}}(aq);\,{{E}^{o}}=-0.71\,V\] \[A{{l}^{3+}}(aq)+3{{e}^{-}}\,\xrightarrow{\,}Al(s);\,\,{{E}^{o}}\,=-1.66\,\text{V}\].

Answer:

                Yes, the tarnish due to silver sulphide will be removed by placing the tarnished ware in an aluminium pan. Actually, aluminium is a stronger reducing agent than silver. Therefore, it will reduce silver ions \[(A{{g}^{+}})\] to metallic silver. It being silvery white in colour will get slowly deposited on the walls of the silver ware and the tarnish will be removed. \[Al(s)+3A{{g}^{+}}(aq)\xrightarrow{\,}\,A{{l}^{3+}}\,(aq)+3Ag(s)\] Alternatively, \[E_{cell}^{o}\,=E_{cathode}^{o}-E_{anode}^{o}=(-0.71)-(-1.66)=0.95\,\text{V}\] is positive. The redox reaction is feasible and the tarnish will be removed.