Based on the cell notation spontaneous reaction- the anode-Ag-s-AgCl-s- Cl - - -aq-parallel - Br - - -aq- Br - 2 -I-C-s-AgCl gets reducedAg gets oxidized- Br - 2 - gets reduced- Br - - - gets oxidized

Question

Toppr Admin · Accepted Answer

Ag-s-AgCl-s-Cl-x2212-x2223-x2223-Br-x2212-x2223-x2223-Br2-C-s-Anode -xA0-Ag-Cl-x2212-x2192-AgCl-e-x2212-Cathode -xA0-12Br2-e-x2212-x2192-Br-x2212-x2234- at anode- oxidation takes place-Hence- Ag gets oxidized-

Based on the cell notation for spontaneous reaction, at the anode:
$A g (s) | A g C l (s) | {C l}^{-} (a q) ∥ {B r}^{-} (a q) | {B r}_{2} (I) | C (s)$
$A g C l$ gets reduced
$A g$ gets oxidized
${B r}_{2}$ gets reduced
${B r}^{-}$ gets oxidized

$A g (s) | A g C l (s) | {C l}^{(-)} ∣ ∣ {B r}^{-} ∣ ∣ | {B r}_{2} | C (s)$
Anode : $A g + {C l}^{(-)} \to A g C l + e^{-}$
Cathode : $\frac{1}{2} {B r}_{2} + e^{(-)} \to {B r}^{(-)}$
$∴$ at anode, oxidation takes place.
Hence, $A g$ gets oxidized.

Based on the cell notation for spontaneous reaction, at the anode:Ag(s)|AgCl(s)|Cl−(aq)∥Br−(aq)|Br2(I)|C(s)AgCl gets reducedAg gets oxidizedBr2 gets reducedBr− gets oxidized

Ag(s)|AgCl(s)|Cl(−)∣∣Br−∣∣|Br2|C(s)Anode : Ag+Cl(−)→AgCl+e−Cathode : 12Br2+e(−)→Br(−)∴ at anode, oxidation takes place.Hence, Ag gets oxidized.

Based on the cell notation for spontaneous reaction, at the anode:
$A g (s) | A g C l (s) | {C l}^{-} (a q) ∥ {B r}^{-} (a q) | {B r}_{2} (I) | C (s)$
$A g C l$ gets reduced
$A g$ gets oxidized
${B r}_{2}$ gets reduced
${B r}^{-}$ gets oxidized

$A g (s) | A g C l (s) | {C l}^{(-)} ∣ ∣ {B r}^{-} ∣ ∣ | {B r}_{2} | C (s)$
Anode : $A g + {C l}^{(-)} \to A g C l + e^{-}$
Cathode : $\frac{1}{2} {B r}_{2} + e^{(-)} \to {B r}^{(-)}$
$∴$ at anode, oxidation takes place.
Hence, $A g$ gets oxidized.