A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+half-cell at 25 C. T

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 1.80 M and 0.120 M , respectively. The volume of half-cells is the same.

1.What is the initial cell potential? ANSWER: 0.56 v

2. What is the cell potential when the concentration of Ni2+ has fallen to 0.600 M ? ANSWER: 0.600 M

3. What is the concentrations of Ni2+ when the cell potential falls to 0.46 V ?

4. What is the concentration of Zn2+ when the cell potential falls to 0.46 V ?

Explanation / Answer

Zn = Zn+2 + 2e- +0.76

Ni+2 + 2e- = Ni -0.25

Zn + Ni+2 == Ni + Zn+2 Eo = 0.51

The number of electrons transferred is 2, so n = 2 in the Nernst Equation

E = Eo - 0.0592/ 2 log [Zn+2] / [Ni+2]

= 0.51 - 0.0296 log(0.120)/(1.40)

= 0.51 - 0.0296(-1.066)

= 0.51 + 0.032

= 0.542

0.46 = 0.51 - 0.0296 log [Zn+2] / [Ni+2]

-0.05 / -0.0296 = log [Zn+2] / [Ni+2]

1.689 = log [Zn+] / [Ni+2]

48.9 = [Zn+2] / [Ni+2]

Let x = change in [Ni+2]

Then [Ni+2] = 1.40 - x

[Zn+2] = 0.120 + x

0.120 + x / 1.40 -x = 48.9

0.120 + x = 68.5 - 48.9x

x = 1.37

[Ni+2] = 1.40 - 1.37 = 0.03

[Zn+2] = 0.129 + 1.37 = 1.499

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.