Question

(Sat)
. The reduction potential of SCE at 25°C is 0.2415 volts, it indicates
(a) Hg2Cl2i) + 2e - → 2Hgo + 2 Clés
(b) 2Hgo + 2 CF (Sat) → Hg2 C1269) + 2e
(c) HgCl2 + 2e → Hg. + 2Cl" (Sat)
(d) Hg** + 2€ Hg(1)​

Answer 1

Answer:

c

Explanation:

i hope my answer help to you

Answer 2

The reduction potential of SCE at 25°C is 0.2415 volts, it indicates

Explanation:

Option (c)

Standard calomel electrode (SCE) : $Hg/Hg_{2}Cl_{2}/KCl$

The reduction potential of SCE is 0.2415 volts at 25°C

The reaction that will take place:

$Hg^{+2}_{2} + 2e^{-} -->2Hg(l)\\Hg^{+2}_{2} + 2Cl^{-} --> Hg_{2}Cl_{2}$

______________________

$Hg_{2}Cl_{2} + 2e^{-}--> 2Hg + 2Cl^{-}$

So, the correct option will be option (c)