# What is the relationship between concentration and voltage in a galvanic cell? How do you calculate the theoretical?

I am a year 12 Chemistry student and I was wondering if you could help me with some questions I have about a Galvanic cell experiment? I am doing an experimental report in Chemistry. The experiment I decided to do was to change the concentration of the copper sulphate electrolyte in a Daniell cell and measure the voltage produced. The other electrolyte; Zinc nitrate was not changed and kept constant at 0.1M. The copper sulphate concentration increased from 0.25M, 0.5M, 0.75M to 1M. The experiment was performed when the temperature was 24 degrees. The anode was Zinc and the cathode was Copper.

In order to obtain theoretical values - should I use the Nernst equation? Would you be able to tell me how to use the Nernst equation for my experiment? Also, I understand that voltage increases with concentration, but compared to standard conditions should the voltage produced be higher or lower than the voltage produced at standard conditions and 1M electrolyte?

• Look at the Nernst equation. You have all the variables. Watch this video. Write a balanced equation, and then follow the instructions youtube.com/watch?v=5Z3SlHgrz0o – M. Farooq Feb 18 at 5:54
• Your questions are answered here: chemistry.stackexchange.com/a/118744/79678 . If you agree, consider upvoting that answer. – Ed V Feb 18 at 16:05
• There has been a publication in Journal of Chemical Education along time ago, saying that zinc should not be used in a zinc nitrate solution, because it slowly reduces the nitrate to nitrite, so that the concentration of zinc is not constant, according to the equation $$\ce{Zn + NO3^- + H2O -> Zn^{2+}.+ NO2^- + 2 OH^-}$$ Better replace zinc nitrate by zinc sulfate ! – Maurice Nov 14 at 16:12

The potential of the Copper electrode is given by the Nernst equation : $$E(Cu^{2+}/Cu) = + 0.34 V + 0.0295 log[Cu^{2+}]$$ If the concentration $$[Cu^{2+}]$$ is $$0.25 M$$ and then $$0.50 M$$, the potential of these half-cells become respectively $$0.32$$ V and $$0.33 V$$.
If you couple this half-cell with a Zinc electrode with $$[Zn^{2+}]$$ = $$0.1 M$$, the same sort of calculation gives you a potential $$E(Zn^{2+}/Zn)$$ =$$-0.76 + 0.0295(-1)$$ = $$-0.79 V$$. If you combine both half-cells, you obtain $$1.11 V$$ and $$1.12 V$$ respectively.