Question

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 1.60molL−1 and 0.130 molL−1, respectively.

Zn2+(aq)+2e−→Zn(s)E∘=−0.76V
Ni2+(aq)+2e−→Ni(s)E∘=−0.23V

Part A

What is the initial cell potential?

Part B

What is the cell potential when the concentration of Ni2+ has fallen to 0.600 molL−1?

Part C

What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?

Answer 1

Answer – Part A) We are given, [Zn²⁺] = 0.130 M , [Ni²⁺] = 1.60 M

Zn²⁺ + 2e^- -----> Zn(s) , E^o = -0.76 V

Ni²⁺ + 2e^- -----> Ni(s) , E^o = -0.23 V

We know the more value of standard reduction potential is act as reduction

Zn(s) -----> Zn²⁺ + 2e^-, E^o = + 0.76 V

Ni²⁺ + 2e^- -----> Ni(s) , E^o = - 0.23 V

Zn(s) + Ni²⁺ ----> Zn²⁺ + Ni(s)

E^ocell = +0.76 -0.23

           = 0.53 V

Now we know the formula for cell potential Nernst equation

Ecell = E^ocell -0.0591/n x log Q

         = 0.53 V – 0.0591 / 2 x log [Zn²⁺] / [Ni²⁺]

         = 0.53 V – 0.0591 / 2 x log 0.130 M / 1.60 M

         = 0.56 V

Initial cell potential is 0.56 V

Part B) Now cell potential when [Ni²⁺] fall to 0.600 M

Now we know formula for cell potential Nernst equation

Ecell = E^ocell -0.0591/n x log Q

         = 0.53 V – 0.0591 / 2 x log [Zn²⁺] / [Ni²⁺]

         = 0.53 V – 0.0591 / 2 x log 0.130 M / 0.60 M

         = 0.55 V

0.55 V cell potential when the concentration of Ni²⁺ has fallen to 0.600 M

Part C) Now we are given cell potential, Ecell = 0.45 V

We know at the equilibrium

    Zn(s) + Ni²⁺ ----> Zn²⁺ + Ni(s)

I             1.60            0.130  

C             -x                 +x   

E        1.60-x             0.130+x

we know,

Ecell = E^ocell -0.0591/n x log Q

     0.45 V = 0.53 V – 0.0591 / 2 x log [Zn²⁺] / [Ni²⁺]

0.45 V – 0.53 V = - 0.0295 x log (0.130+x) / (1.60-x)

log (0.130+x) / (1.60-x) = 2.71

antilog from both side

(0.130+x) / (1.60-x) =509.56

(0.130+x) = (1.60-x) * 509.56

0.130 +x = 815.3 – 509.56x

x + 509.56x = 815.3 -0.130

510.56x = 815.17

x = 1.59

so, [Zn²⁺] = 0.130 +x

                 = 0.130 + 1.59

                 = 1.73 M

[Ni²⁺] = 1.60 –x

          = 1.60 – 1.59

          = 0.0034 M