Answer:
When the nonstandard potential of the cell is 0.02543 v the weight of the cathode is 120.575 g weight.
Explanation:
The half reactions are as follows;
Ni²⁺ + 2 e⁻ → Ni E₀ = -0.231 V
Pb²⁺ + 2 e⁻ → Pb E₀ = -0.133 V
n = 2
Here we have the Nernst equation given by
[tex]E_{cell} = E_{cell}^0 - \frac{RT}{nF} lnQ = E_{cell}^0 - (\frac{0.0591 V}{n} )log Q[/tex]
Where:
[tex]E_{cell}^0[/tex] = 0.098 V
We have, when [tex]E_{cell}[/tex] = 0.02543 V
[tex]E_{cell}[/tex] - [tex]E_{cell}^0[/tex] = 0.02543 - 0.098 = -0.07257 V = [tex]- (\frac{0.0591 V}{2} )log Q[/tex]
log Q = 2.456
Q = [tex]10^{2.456}[/tex] = 285.65
Therefore the number of electron will be
285.65 = the ratio of the Ni²⁺/Pb
Therefore if the initial concentration is 1×10⁻¹ M, we have
[tex]\frac{0.1 + x}{0.1 - x} = 285.65[/tex]
x = 9.93 × 10⁻² Moles
Therefore since 1 mole of Pb = 207.2 g
9.93 × 10⁻² × 207.2 g = 20.575 g
Mass of Pb = 120.575 g.
