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Calculate Ecell for the following electrochemical cell at 25 degreeC Pt(s) | H2(g, 1.00 atm) |...

Calculate Ecell for the following electrochemical cell at 25 degreeC
Pt(s) | H2(g, 1.00 atm) | H+ (aq, 1.00 M) || Pb2+(aq, 0.150 M) | Pb(s)

given the following standard reduction potentials.
   Pb2+(aq) + 2 e– --> Pb(s) Edegree = –0.126 V
      2 H+ (aq) + 2 e– --> H2(g) Edegree = 0.000 V

(Please show the steps)

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Answer #1

E_{cell} = E^o_{Pb} - E^o_{H2} = -0.126 - 0.00 = -0.126 \: V

E_{cell} = E^o_{cell} - \dfrac {0.0592} {n} log \dfrac {[H^+]^2}{[Pb^{2+}]P_{H2}} \\ E_{cell} = -0.126 - \dfrac {0.0592}{2} log \dfrac {1^2}{0.150 \times 1} \\ E_{cell} = -0.126 - 0.0244 = -0.150 \: V

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