A clean metal electrode was placed into each beaker, which was similar to the ion present in the beaker.
The beakers were stuck to the desk with blue-tac.
Crocodile clips and wires were connected a voltmeter between electrode and put into the corresponding beakers, which did not contact with the liquids, and the apparatus was set up as in the diagram below:
Experiment was repeated with the following combinations of metals/ solutions:
-
ZnSO4 & CuSO4 ②FeSO4 & ZnSO4
③NiSO4 & ZnSO4 ④NiSO4 & FeSO4
The results were recorded.
EXPERIMENT III.
Apparatus was set as the following diagram:
A razor blade was used to make cuts in the lemon, before the electrodes were inserted in it.
All results were recorded.
RESULTS
EXPERIMENT I. When closed the switch, the bulb lighted up.
EXPERIMENT II.
① ZnSO4 & CuSO4 : ② FeSO4 & ZnSO4 :
V = 1.108 V V = 0.496 V
Oxidation: Zn(s) Zn2+(a.q) + 2e- Oxidation: Zn(s) Zn2+(a.q) + 2e-
Reduction: Cu2+(a.q) + 2e- Cu (s) Reduction: Fe2+(a.q) + 2e- Fe (s)
Redox equation: Redox equation:
Zn(s)+ Cu2+(a.q) Zn2+(a.q)+ Cu (s) Zn(s)+ Fe2+(a.q) Zn2+(a.q)+ Fe (s)
③ NiSO4 & ZnSO4 ④ NiSO4 & FeSO4
V = 0.956 V V =0. 47 V
Oxidation: Zn(s) Zn2+(a.q) + 2e- Oxidation: Fe(s) Fe2+(a.q) + 2e-
Reduction: Ni2+(a.q) + 2e- Ni (s) Reduction: Ni2+(a.q) + 2e- Ni(s)
Redox reaction: Redox reaction:
Zn(s)+ Ni2+(a.q) Zn2+(a.q)+ Ni (s) Ni(s)+ Fe2+(a.q) Ni2+(a.q)+ Fe (s)
⑤ FeSO4 & CuSO4
V = 0.6V
Oxidation: Fe(s) Fe2+(a.q) + 2e-
Reduction: Cu2+(a.q) + 2e- Cu (s)
Redox reaction:
Fe (s)+ Cu2+(a.q) Fe2+(a.q)+ Cu (s)
EXPERIMENT III.
Zn & Cu: Fe & Zn:
V = 0.839 V V = 0.337 V
Ni & Zn: Ni & Fe:
V = 0.757 V V = 0.513 V
DISCUSS
EXPERIMENT II
A negative EO means that a redox couple is a stronger reducing agent than the H+(a.q)/H2(g) couple.
A positive EO means that a redox couple is a weaker reducing agent than the H+(a.q)/H2(g) couple. [5]
-
ZnSO4 & CuSO4
The Table of Standard Electrode Potentials, which was attached as Appendix A, suggested that Zn is better reducing agent than Cu and Cu2+ ions were better oxygenating agent than Zn2+.
Oxidation = Lose e-
According to fig.1 (page 1, introduction), Zn side was the anode of the cell and Cu side was the cathode of the cell. Cell diagram: Zn(s)| Zn2+(a.q) || Pt | Cu2+(a.q)| Cu (s)
② FeSO4 & ZnSO4 :
According to Appendix A, Zn is better reducing agent than Fe and Fe2+ ions were better oxygenating agent than Zn2+.
Hence, Zn side was the anode of the cell and Fe side was the cathode of the cell.
Cell diagram: Zn(s)| Zn2+(a.q) || Pt | Fe2+(a.q)| Fe (s)
③ NiSO4 & ZnSO4
According to Appendix A, Zn is better reducing agent than Ni and Ni2+ ions were better oxygenating agent than Zn2+.
Hence, Zn side was the anode of the cell and Ni side was the cathode of the cell.
Cell diagram: Zn(s)| Zn2+(a.q) || Pt | Ni2+(a.q)| Ni (s)
④ NiSO4 & FeSO4
According to Appendix A, Fe is better reducing agent than Ni and Ni2+ ions were better oxygenating agent than Fe2+.
Hence, Fe side was the anode of the cell and Ni side was the cathode of the cell.
Cell diagram: Fe(s)| Fe2+(a.q) || Pt | Ni2+(a.q)| Ni (s)
⑤ FeSO4 & CuSO4
According to Appendix A, Fe is better reducing agent than Cu and Cu2+ ions were better oxygenating agent than Fe2+.
Hence, Fe side was the anode of the cell and Cu side was the cathode of the cell.
Cell diagram: Fe(s)| Fe2+(a.q) || Pt | Cu2+(a.q)| Cu (s)
The more negative the EO of a redox couple, the more powerful the reducing agent is of the metal.
∴ the reactivity series of the half cells, from the strongest are: Zn﹥ Fe﹥ Ni﹥ Cu
Prediction of the voltage of a Ni/Ni2+ to Cu/Cu2+ combination:
According to Appendix A, EONi = -0.25 V and EOCu=0.34V
Which the half-reaction were:
Ni(s) Ni2+(a.q)+2e- (-0.25V ) and Cu2+(a.q) + 2e- Cu(s) (0.34V)
Ni(s)| Ni2+(a.q) || Pt | Cu2+(a.q)| Cu (s)
From the formula: EO=EOR-EOL[4]
EO=0.34 – (-0.25) = 0.59 V
∴ The voltage for a Ni/Ni2+ to Cu/Cu2+ combination might be around 0.59 V.
EXPERIMENT III
The lemon battery experiment was the same rule as the experiment II.
Only half lemon was used in this experiment.
Electrodes Zn and Cu were placed into the lemon. Zn was oxidized and ions were coming off. Chemical equation was Zn(s) Zn2+(a.q) + 2e-. The Zn2+ ions were flow into the lemon and Zn became negatively charged due to electron loss.
Cu was reduced by gaining those electrons from Zn via lemon, hence became positively charged. Chemical equation was Cu2+(a.q) + 2e- Cu(s)
Then the ions from negatively charged Zn to positively charged Cu generated electricity and Voltmeter could detect voltage.
Voltmeter could detect voltages among electrodes Fe and Zn, Ni and Zn, and Ni and Fe, but the voltages were different due to the different abilities of reduction and oxidation.
Reference
[1] Lewis R and Evans W(2006) Chemistry. Page100. 3rd Edition. Palgrave Macmillan. New York
[2] Handout from Electrochemistry practical sheet. 29th Apr 2008
[3] Lewis R and Evans W(2006) Chemistry. Page107. 3rd Edition. Palgrave Macmillan. New York
[4] Lewis R and Evans W(2006) Chemistry. Page109. 3rd Edition. Palgrave Macmillan. New York
[5] Lewis R and Evans W(2006) Chemistry. Page110. 3rd Edition. Palgrave Macmillan. New York
