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RYAN INTERNATIONAL SCHOOL,

GHAZIABAD

INVESTIGATORY

PROJECT ON:-

Calculating Gibbs Free energy for redox reactions

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INDEX S no. TOPICS Page no.

1. ACKNOWLEDGEMENT 32. CERTIFICATE 43. INVESTIGATORY PROJECT REPORT

(i) Aim 6(ii) Introduction 6(iii) Materials required 7(iv) Theory 7(v) Procedure 8-9(vi) Observations and Calculations 10(vii) Conclusion 11(viii) References 11

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ACKNOWLEDGEMENT

On completion of my investigatory project I feel extremely delighted but it wasn’t a single effort. First of all I would like to thank my parents for helping me and encouraging me to continue this project and helping me financially as well. Then I would like to thank my teacher Mr. Ranjeet Singh Vyas for teaching me the basics of this investigation. I would also like to thank my chemistry teacher Ms. Preeti Jain for supporting me and giving me the opportunity to make this project. In the end I would like to thank the vegetable vendor, cyber café owner, and Drishti stationeries for providing me with fresh potatoes, printouts and spiral binding respectively.

Shivam Shekhar

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CERTIFICATEThis is hereby to certify that the original and genuine investigation work has been carried out to investigate about the subject matter and the related data collection and investigation has been completed solely, sincerely and satisfactorily by SHIVAM SHEKHAR of CLASS XII – A, Ryan International School, GHAZIABAD,

regarding his project titled “Calculating Gibbs Free energy for redox reactions”.

Teacher’s Signature

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INVESTIGATORY PROJECT ON:-

Calculating Gibbs Free energy for redox reactions

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1. Aim:- To calculate Gibb’s free energy for a redox reaction with the help of electrochemical cell using various electrodes by calculating the value of its EMF.

2. Introduction:- Many chemical reactions we observe around us are basically redox reactions only, occurring in one or the other way. To understand the concepts of redox reactions I am using the electrochemical cell method because it gives a very clear picture of what redox reactions are all about (plus it’s cheap too!!!). Now something about my experiment, I am going to make an electrochemical cell using a potato as electrolyte and metals like iron, Aluminum and copper as electrodes and calculating the value of free energy for a redox reaction by using equation:-

∆G=−nF Ecell

Where; E = EMF of the cell n = no of electrons transferred from anode to cathode in the Reaction equation F= Faraday constant 𝜟G=free energy for redox reaction

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3. Materials Required:-

(i) 1 potato(ii) An aluminum sheet wrapped on a toothpick(iii) An iron nail(iv) A copper coin(v) A multimeter(vi) Space to work

4. Theory:- Potatoes contain several compounds like chlorogenic acid, caffeoylquinic acid, etc. which act as an electrolyte in the experiment. Copper coin, iron nail and aluminum sheet act as electrodes. According to the electrochemical series, amongst iron, copper and aluminum the value of reduction potential are: -

Reaction Eo(Reduction potential)Al3++3e-→Al(s) -1.66 VFe2++2e-→Fe(s) -0.44 VCu2++2e-→Cu(s) 0.34 V

It is clear from the above table that 3 metal couples could be formed i.e. Copper-Aluminum, Copper-Iron, Aluminum-IronNow using equation,

∆G=−nF EcellWe can find the value of E with the help of multimeter, F is already known, n we can get from reaction equation and using all those values we can calculate the value of 𝜟G.

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5. Procedure:- (i) Take the potato and insert the iron nail into it and take a copper

coin and insert it into the potato. (ii) Now take the multimeter and adjust it to measure voltage of the

order of 0 to 2000 mV. The following picture depicts the procedure.

(iii) Fix the multimeter terminals into the potato such that they touch the electrodes and note the reading.

(iv) Repeat the same experiment with Al-Fe and Al-Cu couples and note their readings as well.

Iron nail

Copper coin

Knob at 2000mV position

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With Cu-Al couple:-

With Al-Fe couple:-

Copper coin

Toothpick wrapped with aluminum sheet

Iron nail

Toothpick wrapped with aluminum sheet

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6. Observations and calculations:- From above experiment, we get the following readings:-

Metal couple EMF(in mille volts)Cu-Al 457Cu-Fe 563Al-Fe 398

Now for Cu-Al couple,The reaction is:-2Al→2Al3++6e-

3Cu2++6e-→3Cu2Al+3Cu2+→3Cu+2Al3+

Hence 6e- are transferred in the reaction. Now using the formula we have

∆G=−nF Ecell ∆G=−6∗96500∗0.457 ∆G=−264603 Joules 𝜟G=-264.603 KJSimilarly for other couples,

Couple Reaction EMF 𝜟G=-nFEcell

Cu-Al 3Cu2++2Al→3Cu+2Al3+ 0.457V -264.6KJCu-Fe Cu2++Fe→Cu+Fe2+ 0.563V -108.7KJAl-Fe 2Al+3Fe2+→2Al3++3Fe 0.398V -230.4KJ

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7. Conclusion:- From above experiment it is concluded that free energy change for various metal couples are:-

Couple 𝜟GCu-Al -264.6KJCu-Fe -108.7KJAl-Fe -230.4KJ

8. References:- En.wikipedia.org Google.co.in Chemistry part 2nd class 11th NCERT Chemistry part 1st class 12th NCERT All the pictures have been clicked by Nokia 2700 classic 2 MP

camera with resolution of 1200x1600 All tables along with this project have been created with the help

of Microsoft Word 2007