Based on the cell notation for a spontaneous reaction, at the anode :…
$\mathrm{Ag}(\mathrm{s})|\mathrm{AgCl}(\mathrm{s})| \mathrm{Cl}^{-}(\mathrm{aq}) \| \mathrm{Br}^{-}(\mathrm{aq})\left|\mathrm{Br}_{2}(l)ight| \mathrm{C}(\mathrm{s})$
- AgCl gets reduced
- Ag gets oxidized
- $\mathrm{Br}^{-}$ gets oxidized
- $\mathrm{Br}_{2}$ gets reduced
Solution
Asked in: JEE-TOPICTESTS-CHEMISTRY