Chemistry Forum

Please help me with this task, the result should be -0.366 V. I know how to come up with the result, but in the calculation I took into account that the concentration of sulfate ions is 2 M, which does not seem right to me.

Calculate the potential of the lead electrode in 2 M H₂SO₄ if E0 = -0.126 V and Ksp(PbSO₄) = 1.6 *10-8.

I would like an explanation. Thanks in advance!

Why do you think the concentration of SO₄ 2- is wrong. You have 2 M H₂SO₄ so the value is right.

E =E0 + 0,059V/ z * log c Pb 2+

c Pb 2+ = ksp(PbSO₄) / c SO₄ 2-

cPb 2+ = 1,6 *10^-8 (mol/l)^2 / 2 mol/l = 8 *10^(-9) mol/l

E = -0,126 V+ 0,059/2 * log (8*10^(-9))
E = -0,126V -0,238 V
E = -0,36 V

I understood the task as if at the beginning we only have lead and H₂SO₄ and when they react PbSO₄ is formed, and therefore the concentration of free SO₄ 2- ions must be less than 2 M. But this is obviously not the case and I understand it wrong. Can you please explain to me how to imagine this.

You are right the concentration is less as 2 M
But its 2 mol/l - 8 *10^(-9) mol/l = 1,999999992 mol/l ~ 2 mol/l
So its near the same.