[vectrave1]

Q-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 AlNO3.

Reaction: 2Al(s)+3Pb^(2+) (aq)⟶2Al^(3+) (aq)+3Pb(s)
Electrolytes are AlNO3 and PbNO3.

I am using E_cell=E^0-(R×T)/(n×F)×ln ⁡Q (Nernst equation) and have the following calculation: E_cell=1.53-(8.314×293.7)/(6×96500)×ln⁡〖(.01)^2/(0.1)^3 〗

where Q=(Al^3+)^2/(Pb^2)^3 which is (products)/(reactants)

I am finding that using the nernst equation, an increase in the concentration of AlNO3 leads to a decrease in voltage whereas according to my practical data im finding an increase in voltage.

Also, according to this :https://www.enotes.com/homework-help/galvanic-cell-concentration-anode-solution-307953 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!

[mjc123]

I don't know what that website is but it's talking rubbish. (Basically, it's giving a kinetic answer to a thermodynamic question.) Increasing the concentration of Al³⁺ will decrease the tendency for Al to be oxidised - think Le Chatelier's principle - it will shift the equilibrium to the left. This agrees with what you predict by the Nernst equation (decrease in E_cell = forward reaction less favourable). I don't know why you observe the opposite practically. Is your voltage sensibly close to 1.53V, or is it wildly off?

[chenbeier]

Additionally: An increase of the Al³⁺ will give a higher number in the logarithms term log(c ox /c red.), what also increase the voltage.

[vectrave1]

I don't know what that website is but it's talking rubbish. (Basically, it's giving a kinetic answer to a thermodynamic question.) Increasing the concentration of Al³⁺ will decrease the tendency for Al to be oxidised - think Le Chatelier's principle - it will shift the equilibrium to the left. This agrees with what you predict by the Nernst equation (decrease in E_cell = forward reaction less favourable). I don't know why you observe the opposite practically. Is your voltage sensibly close to 1.53V, or is it wildly off?

Hi,
Yes my reading is wildly off. Keeping the concentration of Pbno3 fixed at 0.1M, for alno3 at 0.01M im getting 0.19v, for 0.1M im getting 0.24 v and at 0.5M im getting 0.29v.

I believe this is due to the oxide layer on aluminum that i didnt remove and the chemicals that were made and then kept in covered flasks for for days before being used

But if you're saying that my nernst readings are correct, what other reasons can i attribute to getting different results practically?

[vectrave1]

Additionally: An increase of the Al³⁺ will give a higher number in the logarithms term log(c ox /c red.), what also increase the voltage.

I'm sorry, i didnt get what you said.
Are you saying that ideally an increase in Alno3 concentration should lead to a higher voltage produced?

[chenbeier]

Yes my reading is wildly off. Keeping the concentration of Pbno3 fixed at 0.1M, for alno3 at 0.01M im getting 0.19v, for 0.1M im getting 0.24 v and at 0.5M im getting 0.29v.

Yes calculate  log (0,01)²/(0,1)³ = -1,  log (0,1)²/(0,1)³ = 1,  log (0,5)²/(0,1)³ = 2,39

So it is an increase .

By the way the formula for Aluminiumnitrate is Al(NO₃)₃ and for lead nitrate Pb(NO₃)₂

[mjc123]

But see the negative sign in the Nernst equation (in this form). It's easy to get this wrong, so I always think Le Chatelier rather than trying to rely on remembering the correct form of the equation.
If your experimental voltages are so far from what they should be, there is something seriously wrong and you can't trust your readings at all. It may be the oxide layer on the aluminium.