Consider the cell :
Pt | H 2 | NaOH (aq), NaCl (aq) | AgCl (s) | Ag
(0.01 M) (0.012 M)
at TºC
E cell = 1.05 V and 
= 0.22 V
Using this knowledge ; and taking 
= 0.06 (log 1.2 = 0.08)
Answer the following questions.
(i) Which of the following is overall cell reaction for the given reaction ?
H 2 (g) + AgCl(s) ⎯→ H + (aq) + Cl¯ (aq) + Ag 14.91 It is greater than 25ºC
Text Solution
Verified by ExpertsA
(i) Overall reaction should be the one which is written in term of species present in the given electrode/cell.
Hence at anode H 2 + 2OH¯ ⎯→ 2H 2 O + 2e –
At cathode 2AgCl + 2e – ⎯→ 2Ag + Cl¯
(ii) We can assume the given cell to be
Pt | H 2 | H + (aq), Cl¯ (aq) | AgCl (s) | Ag
With this assumption , 
= 
– 
= 0.22 V
And cell reaction is
H 2 (g) + AgCl(s) ⎯→ H + (aq) + Ag(s) + Cl¯(aq)
⇒ E cell = 
– 
log (H + ) (Cl – )
1.05 = 0.22 – 0.06 log 
(Cl¯)
0.83 = 0.06 
= pK w – log 
= pK w – log(1.2)
⇒ pK w = 
+ log(1.2) = 13.91
(iii) pK w = 13.91
i.e. K w > 10 –14
Hence T is greater than 25ºC.
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