Chemistry - Electrochemistry Question with Solution | TestHub

In the given cell representation, the left side represents the anode where oxidation takes place, and the right side represents the cathode where reduction takes place.

Oxidation half-reaction at anode:
${\mathrm{A}}_{(\mathrm{s})}\to {\mathrm{A}}_{(\mathrm{aq})}^{+2}+2{\mathrm{e}}^{-}$

Reduction half-reaction at cathode:
${\mathrm{B}}_{(\mathrm{aq})}^{+}+{\mathrm{e}}^{-}\to {\mathrm{B}}_{(\mathrm{s})}$

To balance the number of electrons, multiply the reduction half-reaction by 2:
$2{\mathrm{B}}_{(\mathrm{aq})}^{+}+2{\mathrm{e}}^{-}\to 2{\mathrm{B}}_{(\mathrm{s})}$

Adding the oxidation and reduction half-reactions gives the overall cell reaction:
${\mathrm{A}}_{(\mathrm{s})}+2{\mathrm{B}}_{(\mathrm{aq})}^{+}\to {\mathrm{A}}_{(\mathrm{aq})}^{+2}+2{\mathrm{B}}_{(\mathrm{s})}$

Answer: ${\mathrm{A}}_{(\mathrm{s})}+2{\mathrm{B}}_{(\mathrm{aq})}^{+}\to {\mathrm{A}}_{(\mathrm{aq})}^{+2}+2{\mathrm{B}}_{(\mathrm{s})}$