Question

An electrochemical cell is based on the following two half-reactions:

Ox: Pb(s)→Pb2+(aq, 0.11 M )+2e−

Red: MnO−4(aq, 1.70 M )+4H+(aq, 2.6 M )+3e−→ MnO2(s)+2H2O(l)

Part A Compute the cell potential at 25 ∘C.

Please show all work. Thank you.

Answer 1

Ox:

Pb(s)→Pb2+(aq, 0.11 M )+2e−

Red:

MnO−4 + 4H+ + 3e−→ MnO2(s)+2H2O(l)

Cancel electrons in both the reactions by converting equal number of electrons

So multiply oxidation reaction by 3 and reduction reaction by 2

3Pb(s) → 3Pb2+ + 6e−

2MnO−4 + 8H+ + 6e−→ 2MnO2(s)+4H2O(l)

Summing

3Pb + 2MnO4- + 8H+ -----> 3Pb+2 + 2MnO2 + 4H2O

E = Eo - 0.0591/n *log Q

n = mo of moles of electrons = 6

Q = [Pb+2]^3 / [MnO4-]^2 [ H+]^8

Substitute concentrations

Q = 2.2*10^-7

Eo = E oxidation + E reduction = 0.588 + 0.126 = 0.714 V

Substituting in E cell formula

Ecell = 0.779 V