Respuesta :
Answer :
(a) The initial cell potential is, 0.53 V
(b) The cell potential when the concentration of [tex]Ni^{2+}[/tex] has fallen to 0.500 M is, 0.52 V
(c) The concentrations of [tex]Ni^{2+}[/tex] and [tex]Zn^{2+}[/tex] when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M
Explanation :
The values of standard reduction electrode potential of the cell are:
[tex]E^0_{[Ni^{2+}/Ni]}=-0.23V[/tex]
[tex]E^0_{[Zn^{2+}/Zn]}=-0.76V[/tex]
From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) : [tex]Zn\rightarrow Zn^{2+}+2e^-[/tex] [tex]E^0_{[Zn^{2+}/Zn]}=-0.76V[/tex]
Reaction at cathode (reduction) : [tex]Ni^{2+}+2e^-\rightarrow Ni[/tex] [tex]E^0_{[Ni^{2+}/Ni]}=-0.23V[/tex]
The balanced cell reaction will be,
[tex]Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]
First we have to calculate the standard electrode potential of the cell.
[tex]E^o=E^o_{cathode}-E^o_{anode}[/tex]
[tex]E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}[/tex]
[tex]E^o=(-0.23V)-(-0.76V)=0.53V[/tex]
(a) Now we have to calculate the cell potential.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = emf of the cell = ?
Now put all the given values in the above equation, we get:
[tex]E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}[/tex]
[tex]E_{cell}=0.49V[/tex]
(b) Now we have to calculate the cell potential when the concentration of [tex]Ni^{2+}[/tex] has fallen to 0.500 M.
New concentration of [tex]Ni^{2+}[/tex] = 1.50 - x = 0.500
x = 1 M
New concentration of [tex]Zn^{2+}[/tex] = 0.100 + x = 0.100 + 1 = 1.1 M
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}[/tex]
Now put all the given values in the above equation, we get:
[tex]E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}[/tex]
[tex]E_{cell}=0.52V[/tex]
(c) Now we have to calculate the concentrations of [tex]Ni^{2+}[/tex] and [tex]Zn^{2+}[/tex] when the cell potential falls to 0.45 V.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}[/tex]
Now put all the given values in the above equation, we get:
[tex]0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}[/tex]
[tex]x=1.49M[/tex]
The concentration of [tex]Ni^{2+}[/tex] = 1.50 - x = 1.50 - 1.49 = 0.01 M
The concentration of [tex]Zn^{2+}[/tex] = 0.100 + x = 0.100 + 1.49 = 1.59 M
The cell potential of the given voltaic cell has been 0.49 V. The new cell potential with the decrease in nickel concentration has been 0.52 V. The concentration of nickel has been 0.01 M. The concentration of zinc has been 1.59 M.
What is a voltaic cell?
The voltaic cell has been an electrochemical cell in which the chemical and electrical energy are interconverted. The half cell had the cell potential, and undergoes reduction and oxidation reaction.
The half cell reaction of the voltaic cell has been given as:
[tex]\rm Reduction: Ni^2+\;+\;2e^-\;\rightarrow\;Ni\;E^o=-0.23\\Oxidation: Zn\;\rightarrow\;Zn^2^+\;+\;2e^-\;E^o=-0.76[/tex]
The standard cell potential of the reaction has been given as:
[tex]\rm E^O=E_{Reduction}-E^o_{oxidation}\\E^o=-0.23-(-0.76)\\E^o=0.53\;V[/tex]
- The cell potential can be given as:
[tex]\rm E_{cell}=E^o-\dfrac{0.059}{n}log\dfrac{Zn^2^+}{Ni^2^+}[/tex]
Substituting the values in the above equation:
[tex]\rm E_{cell}=0.53-\dfrac{0.059}{2} log\dfrac{0.1}{1.5}\\ E_{cell}=0.49\;V[/tex]
The cell potential of the given voltaic cell has been 0.49 V.
- The cell potential at the reduced nickel ion concentration to 0.5 M has been:
The concentration of nickel reduced by 0.5, and the concentration of zinc has been added to the reduced.
[tex]\rm Ni^2^+=1.5-x=0.5\\x=1\;M\\Zn^2^+=0.1+x\\Zn^2^+=1.1\;M[/tex]
Substituting the values in Ernst equation:
[tex]\rm E_{cell}=0.53-\dfrac{0.059}{2}log\dfrac{1.1}{0.5} \\E_{cell}=0.52\;V[/tex]
The new cell potential with the decrease in nickel concentration has been 0.52 V.
- The concentration at cell potential 0.45 V can be given as:
[tex]\rm E_{cell}=0.53-\dfrac{0.059}{2}log\dfrac{0.1+x}{1.5-x}\\ x=1.49\;x[/tex]
The concentration of nickel has been:
[tex]\rm Ni^2^+=1.5-x\\Ni^2^+=1.5-1.49\\Ni^2^+=0.01\;M[/tex]
The concentration of nickel has been 0.01 M.
The concentration of zinc has been:
[tex]\rm Zn^2^+=0.1+x\\Zn^2^+=0.1+1.49\\Zn^2^+=1.59\;M[/tex]
The concentration of zinc has been 1.59 M.
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