Answer : The actual cell potential of the cell is 0.47 V
Explanation:
Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
The given redox reaction is :
![Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq%29%2BZn%28s%29%5Crightarrow%20Ni%28s%29%2BZn%5E%7B2%2B%7D%28aq%29)
The balanced two-half reactions will be,
Oxidation half reaction : ![Zn\rightarrow Zn^{2+}+2e^-](https://tex.z-dn.net/?f=Zn%5Crightarrow%20Zn%5E%7B2%2B%7D%2B2e%5E-)
Reduction half reaction : ![Ni^{2+}+2e^-\rightarrow Ni](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%2B2e%5E-%5Crightarrow%20Ni)
The expression for reaction quotient will be :
![Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D%5D%7D)
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
![Q=\frac{(0.0141)}{(0.00104)}=13.6](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%280.0141%29%7D%7B%280.00104%29%7D%3D13.6)
The value of the reaction quotient, Q, for the cell is, 13.6
Now we have to calculate the actual cell potential of the cell.
Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7BRT%7D%7BnF%7D%5Cln%20Q)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 316 K
n = number of electrons in oxidation-reduction reaction = 2 mole
= standard electrode potential of the cell = 0.51 V
= actual cell potential of the cell = ?
Q = reaction quotient = 13.6
Now put all the given values in the above equation, we get:
![E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D0.51-%5Cfrac%7B%288.314%29%5Ctimes%20%28316%29%7D%7B2%5Ctimes%2096500%7D%5Cln%20%2813.6%29)
![E_{cell}=0.47V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D0.47V)
Therefore, the actual cell potential of the cell is 0.47 V