Question

Calculate Ecello. free energy for the following battery. Mg(s) , Mg+2 ; Al+3 , Al(s) Reduction...

Calculate Ecello. free energy for the following battery. Mg(s) , Mg+2 ; Al+3 , Al(s)

Reduction potentials : Al = -1.66V Mg = -2.37 V

Calculate E cell for the new conditons: How does the voltage change and why? (please explain using Le Chateliers principles)

Mg(s) , Mg+2(0.500M) ; Al+3 (2.500M), Al(s)

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

1: Oxidation-half cell: Mg(s) ---- > Mg2+ + 2e- ; E0(oxi) = +2.37 V

Reduction-half cell: Al3+ + 3e- ---- > Al(s); E0(red) = - 1.66

---------------------------------------------------------------------------------------

Overall cell reaction: 3Mg(s) + 2Al3+ ---- > 3Mg2+  + 2Al(s);

E0(cell) = E0(oxi) + E0(red) =   2.37 V - 1.66 V = 0.71 V (answer)

standard free energy, G0 = - nFxE0(cell) = - 6 x 96500 x 0.71 V = 411090 J (answer)

2: The voltage will change according to the concentration of Mg2+ and Al3+ that can be calculated according to Nernst equation.

E(cell) = E0(cell) - (0.0591/n) x log[Mg2+] / [Al3+]

=> E(cell) = 0.71 V - (0.0591/ 6) x log(0.500 M / 2.500 M) = 0.717 V (answer)

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Use the tabulated half-cell potentials to calculate ΔG° for the following redox reaction. 2 Al(s) +...
Use the tabulated half-cell potentials to calculate ΔG° for the following redox reaction. 2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) +6.8 x 102 kJ -7.8 x 102 kJ +4.1 x 102 kJ +1.4 x 102 kJ -2.3 x 102 kJ Answer is 4.1-10^2 kJ Please explain every details When I did it, I got the potential energy -2.37+(-1.66)=-4.03 I don't understand why it's -2.37+1.66!!! Please explain why,
1. Use standard reduction potentials to calculate the standard free energy change in kJ for the...
1. Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 3I2(s) + 2Cr(s) -----> 6I-(aq) + 2Cr3+(aq) Answer: _______ kJ K for this reaction would be greater or less than one. 2. Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2Cu2+(aq) + Sn(s) ----> 2Cu+(aq) + Sn2+(aq) Answer: ______ kJ K for this reaction would be greater or less than one.
1) Fe2+(aq) + 2 Al(s) <==> 2 Al3+(aq) + 3 Fe(s). What is the voltage for...
1) Fe2+(aq) + 2 Al(s) <==> 2 Al3+(aq) + 3 Fe(s). What is the voltage for this cell? 2) What is the Gibbs Free Energy for the above reaction assuming standard conditions? 3. What is the Gibbs Free Energy for the cell in question 1 if the Fe2+ (aq) is [.15] and Al3+ (aq) is [3.80]? 4. Describe what happens within the salt bridge between the two cells. Please show work
3) The corrosion of iron is an electrochemical process that involves the standard reduction potentials given...
3) The corrosion of iron is an electrochemical process that involves the standard reduction potentials given here at 25 °C. Fe2+(aq) + 2e– → Fe(s) E° = –0.44 V O2(g) + 4H+ (aq) + 4e– → 2H2O(l) E° = +1.23 V a. Calculate the voltage for the standard cell based on the corrosion reaction. 2Fe(s) + O2(g) + 4H+ (aq) → 2Fe2+(aq) + 2H2O(l) b. Calculate the voltage if the reaction in Part a occurs at pH = 4.00 but...
± Cell Potential and Free Energy Free-energy change, ΔG∘, is related to cell potential, E∘, by...
± Cell Potential and Free Energy Free-energy change, ΔG∘, is related to cell potential, E∘, by the equation ΔG∘=−nFE∘ where n is the number of moles of electrons transferred and F=96,500C/(mol e−) is the Faraday constant. When E∘ is measured in volts, ΔG∘ must be in joules since 1 J=1 C⋅V. Part A Calculate the standard free-energy change at 25 ∘C for the following reaction: Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) Express your answer to three significant figures and include the appropriate units. Part B...
Short Answer 1) Reduction half-reactions with corresponding standard half-cell potentials are shown below. Zn2+(aq) + 2...
Short Answer 1) Reduction half-reactions with corresponding standard half-cell potentials are shown below. Zn2+(aq) + 2 e‐ → Zn(s) E° = ‐ 0.76 V Al3+(aq) + 3 e‐ → Al(s) E° = ‐ 1.66 V The standard potential for the galvanic cell that uses these two half-reactions is ________ V. 2)In the galvanic cell represented by the shorthand notation shown below, Cd(aq)|Cd2+(aq)|| I2(g)|I-(aq)|Pt(s) the inert electrode is ________, and the balanced cathode half-reaction reaction is ________.
Calculate the Gibbs free energy change at 1200°C for the following reaction: MgS (s) + 3/2...
Calculate the Gibbs free energy change at 1200°C for the following reaction: MgS (s) + 3/2 O2 (g) → SO2 (g) + MgO (s)
What is the standard free energy change and equilibrium constant for the following reaction at 25...
What is the standard free energy change and equilibrium constant for the following reaction at 25 °C? 2Ag+ (aq) + Fe (s)   2 Ag (s) + Fe2+ (aq) Given standard electrode potentials: Ag+ (aq) + e-    Ag (s) E0 = 0.7996 V Fe2+ (aq) + 2 e-    Fe (s) E0 = - 0.447 V
Calculate the standard free energy change, ΔG0r, for the reaction where orthoclase feldspar, KAlSi3O8(s), weathers to...
Calculate the standard free energy change, ΔG0r, for the reaction where orthoclase feldspar, KAlSi3O8(s), weathers to gibbsite, Al(OH)3(s)
For the following reactions and given standard reduction potentials O2(g) + 4H+(aq) + 2Cu(s)  2Cu2+(aq)...
For the following reactions and given standard reduction potentials O2(g) + 4H+(aq) + 2Cu(s)  2Cu2+(aq) + 2H2O(l) O2(g) + 4H+(aq) + 4e-  2H2O(l)       E° = 1.23 V Cu2+ + 2e-  Cu(s)    E° = 0.34 V a. Calculate E°cell b. Calculate ΔG° at 254 K
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT