An electrochemical cell is constructed using half cells in the direction of spontaneous change: Fe(OH)2(s) + 2e- Fe(s) + 2OH-(aq) E = -0.88 V and AgBr(s) + e- Ag(s) + Br-(aq) E = +0.07 V Which of the following options is correct?
- Overall reaction Fe(s) + 2OH-(aq) + 2AgBr(s) Fe(OH)2(s) + 2Ag(s) + 2Br-(aq)
- E cell = -0.95 V
- Fe is reduced in the electrochemical cell
- E cell is an extensive property
Answer
(1) Overall reaction Fe(s) + 2OH-(aq) + 2AgBr(s) Fe(OH)2(s) + 2Ag(s) + 2Br-(aq)
Solution
∵ E (AgBr/Ag) = +0.07 V > E (Fe(OH)2/Fe) = -0.88 V
⇒ AgBr is reduced (cathode) and Fe is oxidised (anode).
Anode: Fe(s) + 2OH-(aq) Fe(OH)2(s) + 2e-
Cathode: 2AgBr(s) + 2e- 2Ag(s) + 2Br-(aq)
Overall: Fe(s) + 2OH-(aq) + 2AgBr(s) Fe(OH)2(s) + 2Ag(s) + 2Br-(aq) ⇒ option (1) is correct.
E cell = E cathode - E anode = 0.07 - (-0.88) = +0.95 V ⇒ option (2) is wrong.
Fe is oxidised, not reduced ⇒ option (3) is wrong.
E cell does not depend on the amount of substance (multiplying the equation does not change it), so it is an intensive property ⇒ option (4) is wrong.
Answer: (1)