Question

Question: Sulfur-containing compounds in the air tarnish silver, giving black Ag₂S . A practical method of cleaning tarnished silverware is to place the tarnished item in electrical contact with a piece of zinc and dip both into water containing a small amount of salt. Write balanced half-equations to represent what takes place

Short answer

Answer

The anode half-equation where oxidation occurs:

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)+2{\text{e}}^{-}$

The cathode half-equation where reduction occurs:

${\text{Ag}}_2{\text{S}}_{\left(s\right)}+{\text{H}}_2{\text{O}}_{\left(l\right)}+2{\text{e}}^{-}\to 2{\text{Ag}}_{\left(s\right)}+\text{H}{{\text{S}}^{-}}_{\left(aq\right)}+{\text{OH}}^{-}\left(aq\right)$

Step-by-step solution

Electrodes

Electrodes are the metal rods used to complete the electrical circuit to pass electricity through an ionic solution. At anode losing of electrons takes place and at cathode gaining of electrons takes place

Effect of air on Sulphur compounds

Compounds which contain sulphur in the air tarnish silver, which give black Ag₂ S.

General information for Anode and Cathode.

To write the half-equations, know some general information:

The significant difference between anode and cathode is that at anode oxidation takes place. As against, at the cathode, reduction occurs.

The anode will attract positively charged particles and the cathode will attract the negatively charged particles.

Balanced half-equations to represent what takes place

Write the anode half-equation where oxidation occurs:

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)+2{\text{e}}^{-}$

The cathode half-equation where reduction occurs:

${\text{Ag}}_2{\text{S}}_{\left(s\right)}+{\text{H}}_2{\text{O}}_{\left(l\right)}+2{\text{e}}^{-}\to 2{\text{Ag}}_{\left(s\right)}+\text{H}{{\text{S}}^{-}}_{\left(aq\right)}+{\text{OH}}^{-}\left(aq\right)$