Question

In: Chemistry

An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.19 M )+2e− Red:MnO−4(aq,...

An electrochemical cell is based on the following two half-reactions:
Ox: Pb(s)→Pb2+(aq, 0.19 M )+2e−
Red:MnO−4(aq, 1.65 M )+4H+(aq, 1.9 M )+3e−→
MnO2(s)+2H2O(l)

Compute the cell potential at 25 ∘C.

Expert Solution

Pb(s)→Pb2+

MnO−4(aq, 1.65 M )+4H+(aq, 1.9 M )+3e−

The euqations:

MnO4− + 4 H+ + 3 e− <-----> MnO2(s) + 2 H2O +1.70

Pb2+ + 2 e− <-----> Pb(s) −0.126

most negativ eiwll oxidize

MnO4− + 4 H+ + 3 e− <-----> MnO2(s) + 2 H2O +1.70

Pb(s) <-----> Pb2+ + 2 e− +0.126

E°cell = 1.70+0.126 = 1.826 V

For n and Q we need the equaiton

MnO4− + 4 H+ + 3 e− <-----> MnO2(s) + 2 H2O

Pb(s) <-----> Pb2+ + 2 e− +0.126

Balance electrons

2MnO4− + 8H+ + 6e− <-----> 2MnO2(s) + 4H2O

3Pb(s) <-----> 3Pb2+ + 6e−

Add both equations

2MnO4− + 8H+ + 6e− + 3Pb(s) <-----> MnO2(s) + 4H2O + 3Pb2+ + 6e−

Cancel common terms

2MnO4− + 8H+ + 3Pb(s) <-----> MnO2(s) + 4H2O + 3Pb2+

Then

n = 2*3 = 6

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

Q = (0.19^3) / ((1.65^2)*(1.9^8)) = 0.0000148

substitute in nerst equation

Ecell = E°Cell - 0.0592/n*log(Q)

Ecell = 1.826 - 0.0592/6*log(0.0000148) = 1.87365

queen_honey_blossom answered 1 hour ago

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