A) What is the initial cell potential? A voltaic cell consists of a Zn/Zn^2+ hal

Question

A) What is the initial cell potential? A voltaic cell consists of a Zn/Zn^2+ half-cell and a Ni/Ni^2+ half-cell at 25C . The initial concentrations of Ni^2+ and Zn^2+ are 1.50M and 0.130M , respectively. The volume of half-cells is the same. What is the initial cell potential? What is the cell potential when the concentration of Ni^2+ has fallen to 0.600M ? What is the concentrations of Ni 2+ when the cell potential falls to 0.46V ? What is the concentration of Zn 2+ when the cell potential falls to 0.46V ?

Explanation / Answer

The reaction will be as follows :
Zn + Ni2+ --> Zn2+ + Ni

Oxidation Reaction :

Zn --> Zn2+ + 2e-
Eo oxid = 0.7628

Reduction Reaction :
Ni2+ + 2e- --> Ni
Eo red = -0.23

(a)

Initial Cell potential :

= 0.7628 - 0.23

= 0.5328 volts

(b)
Initial Concentration of Ni+2 = 1.50
And we want to calculate when it decreases to 0.600
Change = 1.5 - 0.600 = 0.900 M
Now conc of Zn+2 should also be increased by same amount from 0.100 M to 1 M(0.100 + 0.900).

From Nernst Equation :
Q = [Zn2+]/[Ni2+]
= 1 /0.600
= 1.67

Ecell = E0cell - (RT/nF)lnQ

At room temperature (25

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