# Concentration Cell

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#### DeathandTaxes

##### Full Member
7+ Year Member For question 79, how are we supposed to know that Q is Anode/Cathode instead of the other way around? Also, why is the standard E set to 0?

#### Czarcasm

##### Hakuna matata, no worries.
5+ Year Member
View attachment 181589

For question 79, how are we supposed to know that Q is Anode/Cathode instead of the other way around? Also, why is the standard E set to 0?
Hmm. Electrochemistry Kill me now, lol. I'm still learning this topic so take this with a grain of salt. Looking at the diagram, we have two cells, both with the same metal and metal ion but in unequal concentrations. Glancing at the table of half-reactions, we see that Cu2+ reduction potential is spontaneous (a positive value). Cu2+ wants to gain electrons. Considering we have 1000 times more Cu2+ in cell 2 relative to cell 1, this reaction will occur spontaneously to consume some of that excess Cu2+ until we establish equilibrium (at which point there is no net consumption of e-). And because we now realize that cell 2 is gaining electrons (being reduced), we know this is our anode (and cell 1 is our cathode).

In the nernst equation, the Q component is Products over Reactants. The only species that will appear in this expression are the aqueous ions (not Cu(s)). We have 0.001 Cu2+ Products and 1M Cu2+ Reactants. 1:1000 ratio which represents a log value of -3. Our 'n' value are the moles of electrons being consumed in the half reaction (2).

And our E cell standard for the half reaction is: 0.34 V (provided by the table).
Correction: E cell standard is total value of the E cell standard for the reduction half reaction and oxidation half reaction. Because these values are equal and opposite, the total value for E cell standard is zero. (What I basically did was: E cell standard = Ered but didn't add in Eox); This is actually a unique feature of concentration cells (E cell standard overall is 0V).

Then it becomes plug and chug:

0 - (-3)(0.06)/2 ==> 0 + 0.18/2 ==> 0 + 0.9 ==> 0.9 V equals E non-standard.

Disregard this, made a mistake: 0.34 - (-3)(0.06)/2 ==> 0.34 + 0.18/2 ==> 0.34 + 0.9 ==> 0.43 V equals E non-standard.

Does that make intuitive sense with what we predicted? Well yes, it's a positive value (indicating it's spontaneous) and as we predicted, some of the 1M Cu2+ would be consumed to establish equilibrium. Seems correct to me.

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#### Czarcasm

##### Hakuna matata, no worries.
5+ Year Member
EDIT: Corrected mistake in above post.

I made a mistake actually. E cell standard is zero (E cell standard: Ean + Ecat) And because the two reactions are equal and opposite in value, the overall E cell standard is therefore zero. (I included only the reduction half reaction as my value for Ecell standard). So the correct answer actually should be B, 0.089.

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#### DeathandTaxes

##### Full Member
7+ Year Member
Cool, thanks!

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