CHEMISTRY HIGH SCHOOL

77. A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25°C. The initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M, respectively. a. What is the initial cell potential? b. What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M? c. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?

Answers

Answer 1

Answer:

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^(2+)$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^(2+)$ and $Zn^(2+)$ 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:

$E^0_([Ni^(2+)/Ni])=-0.23V$

$E^0_([Zn^(2+)/Zn])=-0.76V$

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) : $Zn\rightarrow Zn^(2+)+2e^-$ $E^0_([Zn^(2+)/Zn])=-0.76V$

Reaction at cathode (reduction) : $Ni^(2+)+2e^-\rightarrow Ni$ $E^0_([Ni^(2+)/Ni])=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^(2+)(aq)\rightarrow Zn^(2+)(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_(cathode)-E^o_(anode)$

$E^o=E^o_([Ni^(2+)/Ni])-E^o_([Zn^(2+)/Zn])$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_(cell)$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_(cell)=0.53-(0.0592)/(2)\log ((0.100))/((1.50))$

$E_(cell)=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^(2+)$ has fallen to 0.500 M.

New concentration of $Ni^(2+)$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^(2+)$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])$

Now put all the given values in the above equation, we get:

$E_(cell)=0.53-(0.0592)/(2)\log ((1.1))/((0.500))$

$E_(cell)=0.52V$

(c) Now we have to calculate the concentrations of $Ni^(2+)$ and $Zn^(2+)$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)+x])/([Ni^(2+)-x])$

Now put all the given values in the above equation, we get:

$0.45=0.53-(0.0592)/(2)\log ((0.100+x))/((1.50-x))$

$x=1.49M$

The concentration of $Ni^(2+)$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^(2+)$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

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Answers

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Answers

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Answers

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Answers

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