Question

Calculate the cell potential for the reaction as written at 25.00 °C, given that [Zn2+] = 0.839 M and [Ni2+] = 0.0150 M. Use the standard reduction potentials in this table. Zn(s) + Ni2+(aq) = Zn2+ (aq) + Ni(s) E = V

Answer 1

To calculate the cell potential we have the relation called nearst equation,

$E_{cell}=E_{cell}^{o}-\frac{RT}{nF}ln(Q)$

where E^o_cell= standard cell potential,

R= gas constant= 8.314 J/mol-K

T= 25^oC = 298K ,

n= number of electrons involved

from the reaction we can see that zinc is losing 2 elctrons and nickel is gaining 2 electrons

so, n=2

F = faraday constant= 96500

Q= Reaction quotient= [Product]^a/[Reactant]^b= concentration of product raised to the power its coefficient/concentration of reactant raised to the power its coeffeicient

reduction potential of zinc= -0.76V

reduction potential of nickel= -0.26V

E^o_cell= E^o_cathode- E^o_anode

anode is that one where oxidation takes place(ZINC)

cathode is where reduction takes place(NICKEL)

E^o_cell= -0.26-(-0.76)

E^o_cell= -0.26+0.76

E^o_cell= 0.50V

Q= [Product]^a/[Reactant]^b

Q= [Zn²⁺]/[Ni²⁺]

Q= 0.839/0.0150

Q= 55.93

$E_{cell}=E_{cell}^{o}-\frac{RT}{nF}ln(Q)$

Putting all the values,

$E_{cell}=0.50-\frac{8.314\times 298}{2\times 96500}ln(55.93)$  

E_cell= 0.5-0.0128.ln(55.93)

E_cell= = 0.5-0.0515

E_cell= 0.448V

the cell potential for the reaction is 0.448V

I hope this helps you..