Question

For the galvanic cell 

Ag | AgCl_(s) KCl (0.2M) || KBr (0.001 M) AgBr_{(s) |} Ag 

Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at 25°C

[K_sp(AgCl) = 2.8 x 10^–10; K_sp (AgBr) = 3.3 x 10^–13]

Answer 1

The given cell may be written as under 

Ag|Ag⁺(C₁)Cl^– (0.2M) || Ag⁺ (C₂), Br^– (0.001 M) | Ag 

The above cell in a concentration cell with the cell reaction[ Ag⁺] _RSH → [Ag ⁺] _LHS

= – 0.0591 log 4.242 = – 0.0591 x 0.6276 = – 0.03709 volts 

Since cell potential is negative, the cell reaction will be spontaneous in the reverse direction i.e. AgCl + Br^– → AgBr + Cl^– 

with Cl^– | AgCl | Ag serving as cathode (+ve terminal) and Br^– | AgBr | Ag as anode ( –ve terminal). The spontaneity is due to the fact that AgBr is less soluble than AgCl.