6N.7.b & n in nernst equation

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6N.7.b & n in nernst equation

Postby 205405339 » Tue Mar 03, 2020 9:41 pm

so i thought n was the moles of electrons transferred but thats not the case for 6N.7.b. can someone explain

Julie Park 1G
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Re: 6N.7.b & n in nernst equation

Postby Julie Park 1G » Tue Mar 03, 2020 10:19 pm

The reduction half-reaction involved is 2H+ + 2e- --> H2
And the final cell reaction is described by H+ (aq, 1x10-3M) --> H+ (aq, 1x10^-4M)

When you add the reduction half-reactions together (by reversing one of the equations), you will get
2H(aq, 1x10^-3M) --> 2H+ (aq, 1x10^-4M)
I wonder if they used 1 as the # of moles of electrons transferred by considering that the final, simplified equation involves only 1mol of reactants and products instead of two (as presented in the equation above).

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