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The emf of Cd|Cd2+ (? M) || Ni2+ (1.0 M) | Ni is 0.24 V. What...

The emf of Cd|Cd2+ (? M) || Ni2+ (1.0 M) | Ni is 0.24 V. What is the concentration of Cd2+?

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Answer #1

Lets find Eo 1st

from data table:

Eo(Cd2+/Cd(s)) = -0.4 V

Eo(Ni2+/Ni(s)) = -0.25 V

As per given reaction/cell notation,

cathode is (Ni2+/Ni(s))

anode is (Cd2+/Cd(s))

Eocell = Eocathode - Eoanode

= (-0.25) - (-0.4)

= 0.15 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

we have below equation to be used:

E = Eo - (0.0592/n) log {[Cd2+]^1/[Ni2+]^1}

0.24 = 0.15 - (0.0592/2) log ([Cd2+]^1/1.0^1)

log ([Cd2+]^1/1.0^1) = -3.041

([Cd2+]^1/1.0^1) = 9.109*10^-4

[Cd2+] = 9.11*10^-4 M

Answer: 9.11*10^-4 M

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