![13. What is the electric potential (E) for the following redox reaction at 298K when [Ni²*] = 3.0 M,
and [Zn*] = 0.1 M?
Zn
Zn²*
Ni](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F319348e8-ec3b-434b-beac-e2801422c409%2F17db2a63-1191-4fa7-96ed-e1e7a619f08a%2Fa4pbbg_processed.png&w=3840&q=75)
The balanced reaction taking place is given as,
=> Zn + Ni2+ ---------> Ni + Zn2+
Given : [Ni2+ ] = 3.0 M
And [Zn2+ ] = 0.1 M
In the above reaction, the oxidation state of Zn is increasing from 0 to 2 and Ni is decreasing from 2 to 0.
Hence the reduction of Ni and oxidation of Zn is taking place in the above reaction.
Hence the standard cell potential of the above reaction is given by,
=> Eocell = Ered - Eox
where Ered = reduction potential of species being reduced = -0.28 V
And Eox = reduction potential of species being oxidised = -0.76 V
Hence Eocell = -0.28 - (-0.76) = 0.48 V
Also, in the above reaction, 1 Zn is gaining 2 electrons to become Zn2+.
Hence the number of electrons transferred in the reaction = n = 2.
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