6th edition, 14.37

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Hai-Lin Yeh 1J
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6th edition, 14.37

Postby Hai-Lin Yeh 1J » Thu Mar 07, 2019 3:15 pm

Determine the potential of each of the following cells:
(a) Pt(s) | H2(g, 1.0 bar) | HCl(aq, 0.075 m) || HCl(aq, 1.0 mol.L.-1) | H2(g, 1.0 bar) | Pt(s)

In the solutions manual, they got 2H+ + 2e- -> H2 (g) as the equation for the cathode and H2 (g) -> 2H+ (aq) + 2e- as the anode. How did that happen when there is an HCl on both sides? Where did the Cl go?

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Re: 6th edition, 14.37

Postby Chem_Mod » Thu Mar 07, 2019 4:40 pm

It is a concentration cell and the Cl- is free to move around between anode and cathode.

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