A galvanic cell is set up with a copper electrode in contact with 0.1M CuSO4(aq) and a lead electrode in contact with 0.1M Pb(NO3)2(aq), respectively, at 25ºC.
The standard reduction potentials are:
Pb2+ + 2 e- → Pb; Eº= -0.13 V
Cu2+ + 2 e- → Cu; Eº= +0.34 V
If H2SO4 is being added to the half-cell containing Pb(NO3)2 (thus forming a precipitate of PbSO4), until the final concentration of SO2- 4ions in solution is 2.4 x 10-1 M, what is the cell potential in Volt? Assume the volume of the the half-cell does not change upon addition of H2SO4. Ksp(PbSO4)=6.3 x10-7?
The cell reaction is Cu²⁺ + Pb(s) ⇌ Cu(s) + Pb²⁺
E°(cell) = 0.34 V + 0.13 V = +0.47 V
Now we add SO₄²⁻ and set up the solubility equilibrium until [SO₄²⁻] = 2.4 × 10⁻¹
PbSO₄(s) ⇌ Pb²⁺ + SO₄²⁻ Ksp = 6.3 × 10⁻⁷ = [Pb²⁺][SO₄²⁻]
then [Pb²⁺] = (6.3 × 10⁻⁷)/(2.4 × 10⁻¹) = 2.625 × 10⁻⁶
The +0.47 V cell voltage only applies to the standard cell, where the concentration of each specie = 1.0 M.
To get the voltage for the concentrations we actually have, use the Nernst equation:
E(cell) = E°(cell) – () ln
= E°(cell) – () log₁₀
for T = 298 K
n = no of moles of electrons transferred in reaction = 2
[Pb²⁺]/[Cu²⁺] = 2.625 × 10⁻⁵
E(cell) = +0.47 – (0.0592/2) log₁₀ (2.625 × 10⁻⁵)
= 0.47 + 0.136 V = 0.606 V
E°(cell) = 0.34 V + 0.13 V = +0.47 V
Now we add SO₄²⁻ and set up the solubility equilibrium until [SO₄²⁻] = 2.4 × 10⁻¹
PbSO₄(s) ⇌ Pb²⁺ + SO₄²⁻ Ksp = 6.3 × 10⁻⁷ = [Pb²⁺][SO₄²⁻]
then [Pb²⁺] = (6.3 × 10⁻⁷)/(2.4 × 10⁻¹) = 2.625 × 10⁻⁶
The +0.47 V cell voltage only applies to the standard cell, where the concentration of each specie = 1.0 M.
To get the voltage for the concentrations we actually have, use the Nernst equation:
E(cell) = E°(cell) – () ln
= E°(cell) – () log₁₀
for T = 298 K
n = no of moles of electrons transferred in reaction = 2
[Pb²⁺]/[Cu²⁺] = 2.625 × 10⁻⁵
E(cell) = +0.47 – (0.0592/2) log₁₀ (2.625 × 10⁻⁵)
= 0.47 + 0.136 V = 0.606 V