# Does increasing the concentration of the anode electrolyte in a galvanic cell increase or decrease the voltage produced

Im in kind of a pickle here. I am using nernst equations to calculate the theoretical voltage produced for different concentrations of $$\ce{Al(NO3)3}$$.

Reaction: $$\ce{2Al(s) + 3Pb^{2+}(aq)⟶2Al^{3+}(aq) + 3Pb(s)}$$

Electrolytes are $$\ce{Al(NO3)3}$$ and $$\ce{Pb(NO3)2}$$.

I am using the Nernst equation $$E_{cell}=E^0-\frac{RT}{nF}\mathrm{ln}⁡(⁡Q)$$ and have the following calculation: $$E_{cell}=1.53-\frac{(8.314×293.7)}{(6×96500)}×\mathrm{ln}⁡\left(\frac{(.01)^2}{(0.1)^3} \right)$$

where $$Q=[Al^{3+}]^2/[Pb^{2+}]^3$$ is the ratio (products)/(reactants).

I am finding that using the Nernst equation, an increase in the concentration of $$\ce{Al(NO3)3}$$ leads to a decrease in voltage whereas according to my practical data im finding an increase in voltage.

Also, according to this an increase in concentration should lead to an increase in voltage.

Can anybody tell me what im doing wrong here.

Any help would be appreciated!

• Well, the linked answer is completely wrong. Since this is an equilibrium process, an increase in the concentration of the product will make the reaction less spontaneous. In a galvanic cell, this translates to a smaller potential difference. – Zhe Apr 2 '19 at 20:39
• I've never heard of eNotes before, but judging by this example, what you did wrong was to trust the content on that site. – Zhe Apr 2 '19 at 20:41