In the cell reaction
`Ag_((s))+Cu(aq))^(2+)+Br_((aq))^(-) to AgBr_((s))+Cu_((s))`
the reduction half reaction is

Options

A. `Cu +2e^(-) rarr Cu^(2-)`
B. `Cu - 2e^(-) rarr Cu^(2+)`
C. `Ag^(+) + e^(-) rarr Ag`
D. `Ag- e^(-) rarr Ag^(+)`

Answer

C

Explanation

`Cu(S) + 2Ag^(+)(aq) rarr Cu^(2+)(aq) + 2Ag(s)`
`Ag^(+)(aq)` is reducing to Ag(s)
`Ag^(+)(aq) + e^(-) rarr Ag(s)`