1) Using the standard reduction potentials table in your book/notes, what is the

Question

1) Using the standard reduction potentials table in your book/notes, what is the standard change in gibbs free energy for the following reaction in kJ?

2NO3-(aq) + 8H+(aq) + 3Cu(s) 3Cu2+(aq) + 2NO(g) + 4H2O(l)

If you took CHEM 122 lab here, this was the first reaction in the cycle of copper lab that released the brown NO gas.

2) What is the value of K at 298K for the following reaction? (Please give your answer to two significant digits in scientific notation) K = enFE°/RT

3A2+(aq) + 2B(s) 2B3+(aq) + 3A(s)

A2+(aq) + 2e- A(s) E°red = 0.58 V

B3+(aq) + 3e- B(s)  E°red = 0.75 V

3) What is the cell potential for the spontaneous concentration cell with the reaction Fe3+(aq) + Fe(s) Fe(s) + Fe3+(aq), with concentrations of Fe3+ of 0.848M and 0.353M at a temperature of 310. K?

Ecell = E°cell (RT/nF)lnQ   

4) Which of the following will increase the cell potential for the following reaction:

2Fe3+(aq) + 3Zn(s) 3Zn2+(aq) + 2Fe(s)

Use the reaction Ecell = E°cell (RT/nF)lnQ and consider how each manipulation affects Q and thus affects Ecell

- Add zinc solid

- Remove water from the entire system

- Add solid iron to the system

- Increase the concentration of zinc ion in solution

- Dilute the entire system

- Increase the concentration of Fe3+ in the solution

Explanation / Answer

Hi, according to Chegg question policy, I will be answering the first question for you.

1.

First, we will write separate reduction/ oxidation reactions for the redox reaction:

Anode (oxidation) : Cu --> Cu2+ + 2e ...Equation 1 (EoCu2+|Cu = 0.337V)

Cathode (reduction) : NO3- + 4H+ + 3e --> NO + 2H2O ...Equation 2 (EoNO3-|NO= 0.958V)

Multiplying equation 1 by 3 and equation 2 by 3, to equalize the number of electrons on both sides, and adding the two equations,

2NO3- + 8H+ + 3Cu +6e 3Cu2+ + 2NO + 4H2O + 6e

Now, the standard cell potential will be given by-

Eocell = Eocathode - Eoanode

= EoNO3-|NO - EoCu2+|Cu = 0.958 - 0.337 = 0.621 V

Now, the standard change in Gibbs free energy is given by:

Go = -nFEocell

where n= number of electrons involved in the redox reaction (6 in this case)

F = Faraday's constant = 96500 C

So, Go = - 6 X 96500 X 0.621 V

= 359559 J or 359.559 kJ

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