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factors affecting voltage of voltaic cell

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1. I have to design a lab about the factors affecting voltage of a voltaic cell. I was thinking of different surface areas of the electrodes... zinc and copper, I guess. I'll keep copper constant and change the surface area of zinc. But what sort of effect would this have on the voltage?

I was also thinking of changing the temperature - keeping copper constant and changing temperature of zinc, but what effect would this have?

THANKS.
2. I think the only factor affecting the voltage is the type of material used for the electrodes. Voltage is the potential difference between the two electrodes - if there are more ions moving, or they're moving faster, then only the current increases, not the voltage.
3. (Original post by Zygroth)
I think the only factor affecting the voltage is the type of material used for the electrodes. Voltage is the potential difference between the two electrodes - if there are more ions moving, or they're moving faster, then only the current increases, not the voltage.
Not so, look up Nernst equation for example.

The Eº values apply only to standard conditions.
4. Didn't think the surface area would affect it Vary the concentration and temperature of your electodes. Your cell reaction will be:
Zn2+(aq) + 2e- ↔ Zn(s) Eº = -0.763V
Cu2+(aq) + 2e- ↔ Cu(s) Eº = +0.340V
Cu2+(aq) + Zn(s) ↔ Cu(s) + Zn2+(aq) Eº = +1.103V

You can write a Nernst equation for this, or Nernst equations for each half cell reaction. You can then see how your varying the concentration and temperature will affect this.
5. when the voltage is positive, is it on the left side or right side?
IB doesnt teach Nernst equation >.<
anyway to explain it without using Nernst equation?
6. When it's positive it goes from left -> right, i.e. the reaction is "feasible". Nernst might appear confusing at first but once you understand it you should be okay .

- E = voltage/potential (V) that you'll be measuring
- Eº = standard electrode potential (V). For the above reaction it's +1.103V, treat it as a constant for the cell reaction you do. It will change if you change the electrodes, but it won't change if you have the same electrodes.
- R = gas constant = 8.314 J K−1 mol−1
- T = temperature (K)
- n = no of electrons transferred. Look at the balanced half-equations and see how many electrons are transferred, e.g. for the above it's 2
- F = Faraday constant = 9.648×104 C mol−1
- Q = reduction/oxidation. Zinc and copper are both solids, so you can ignore them. Q will therefore = [Cu2+]/[Zn2+]; the concentration of the Cu2+ and Zn2+ ions.

The variables are temperature and concentration of Cu2+/Zn2+, your dependent variable will be potential.

An increase in temperature depends on your concentration of Cu2+ and Zn2+ because that affects the sign of ln. If your Cu2+ concentration is smaller than your Zn2+ concentration, an increase in temperature will result in an increase in potential. If your Cu2+ concentration is bigger than your Zn2+ concentration, an increase in temperature will result in an decrease in potential (i.e. the reverse).

Now lets say you keep the concentration of Zn2+ the same, but increase the concentration of Cu2+, you get a decrease in potential. If you keep the concentration of Cu2+ the same and increase the concentration of Zn2+, you get an increase in potential.

That might appear really confusing so ask as many questions as you want . It helps to plug values into the equation, change them around and see the effect.
7. i am so confused with your example. i changed the concentration of a voltaic cell. i changed the concentration of ZnSO4, and left the concentration the same for the CuSO4. Can you help me to figure out Nernst's equation using my data?
8. okay, since Zn is more reactive than Cu, then it has more tendancy to lose electron and become a positive ion, then the Zn equation is " Zn ==> (reversible) (Zn+2) + (2e) "
and for the CU since to was reduced, its equation will be
" (Cu+2) + (2e) ==> (reversible) Cu" ;]
9. (Original post by ChemicalQ)
okay, since Zn is more reactive than Cu, then it has more tendancy to lose electron and become a positive ion, then the Zn equation is " Zn ==> (reversible) (Zn+2) + (2e) "
and for the CU since to was reduced, its equation will be
" (Cu+2) + (2e) ==> (reversible) Cu" ;]

While I'm sure that your contribution is well-meaning, I think you'll find that the thread is about 18 months old...

Updated: September 28, 2010
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