6N3 a) Concentrations?

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6N3 a) Concentrations?

Postby nicolely2F » Tue Mar 03, 2020 4:48 pm

Predict the potential of each of the following cells:
(a) Pt(s)|H2(g, 1.0 bar)|HCl(aq, 0.075 mol/L) || HCl(aq, 1.0 mol/L)|H2(g, 1.0 bar)|Pt(s)
The solutions manual shows ln Q = (0.075^2 / 1.0^2).
Why is 0.075 M plugged in the place of the product and 1.0 M (cathode) plugged in the place of the reactant in ln Q? I thought it'd be the other way around since electrons flow from anode to cathode.

Alicia Lin 2F
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Re: 6N3 a) Concentrations?

Postby Alicia Lin 2F » Tue Mar 03, 2020 5:38 pm

You would look at the half reactions. At the anode, it is H2-->2H+ +2e-. At the cathode, it is 2H+ +2e- -->H2. Q is always product over reactant. So you take the products of both equations and multiply those values as the numerator. Then you take the reactants of both equations and multiply those values as the denominator. 0.075M is on top because it is the product of the oxidation reaction at the anode.

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