July 17, 2024, 06:28:22 PM
Forum Rules: Read This Before Posting


Topic: Computation of emf of the reaction in non-standard conditions  (Read 2065 times)

0 Members and 1 Guest are viewing this topic.

Offline Win,odd Dhamnekar

  • Full Member
  • ****
  • Posts: 165
  • Mole Snacks: +0/-5
  • Gender: Male
  • Stock Exchange Trader, Investor,Chemistry Hobbyist
Computation of emf of the reaction in non-standard conditions
« on: August 28, 2022, 11:13:00 AM »
Be sure to answer all parts.

Consider the following reaction:

Cd(s) + Fe2+(aq) :rarrow: Cd2+ + Fe(s)

E°(Fe2+/Fe)= -0.4400 V, E°(Cd2+/Cd)= -0.4000 V.

Compute the emf for this reaction at 25°C if [Fe]2+= 0.75 M, [Cd]2+= 0.010 M.

Will the reaction occur spontaneously at these conditions?

My answer: E°Fe(cathode)(+)- E°Cd(anode)(-)= [itex]\frac{0.059}{2}\log{\frac{[0.01]}{[0.75]}} = -0.06V[/itex]
-0.4400 V-(-0.06)=-0.3847 V

ΔG°= -2× 96485.3399 J/(V-mol) × -0.3847 V= 7 × 101 kJ/mol

As ΔG° is + ve, this reaction is non-spontaneous.

Are my above answers correct? Are my unit conversions and computation of emf of this reaction correct? 

Any science consists of the following process.
 1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference 8) take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2068
  • Mole Snacks: +302/-12
Re: Computation of emf of the reaction in non-standard conditions
« Reply #1 on: August 28, 2022, 11:53:35 AM »
No and no. What formula are you trying to use? Are you understanding the difference between E and E°?

Offline Win,odd Dhamnekar

  • Full Member
  • ****
  • Posts: 165
  • Mole Snacks: +0/-5
  • Gender: Male
  • Stock Exchange Trader, Investor,Chemistry Hobbyist
Re: Computation of emf of the reaction in non-standard conditions
« Reply #2 on: August 28, 2022, 01:07:36 PM »
 I think my answers are wrong.
My corrected answers:
 
For the half reaction(oxidation) Cd(s)  :rarrow: Cd2+(aq) + 2 e-, E° = + 0.4000 V

The nernst expression is E = E° - 0.059/n log ([reduction]b/ [ox]a)
 [itex]E= -0.4000 V + \frac{0.059}{2}\log{\left(0.010\right)}= -0.46 V[/itex]
 
For the half reaction(reduction) Fe2+(aq) + 2 e- :rarrow: Fe(s), E° = -0.4400 V

[itex] E= -0.4400 V + \frac{0.059}{2}\log{\left(\frac{1}{[0.75]}\right)} = -0.444 V [/itex]
 
E for the complete reaction Ereduction(cathode)+- Eox(anode)-= -0.444 V -(-0.46 V)= 0.02 V

As E° for the whole reaction is +ve, we can say this reaction is spontaneous.

Are these above answers correct?
« Last Edit: August 28, 2022, 01:49:03 PM by Win,odd Dhamnekar »
Any science consists of the following process.
 1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference 8) take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5651
  • Mole Snacks: +327/-24
Re: Computation of emf of the reaction in non-standard conditions
« Reply #3 on: August 29, 2022, 11:26:34 AM »
Terminology may not be uniform among all textbooks and teachers.  I prefer to use spontaneous and non-spontaneous for ΔG and to use favorable and unfavorable for ΔG°.  My suggestion to you is to follow the the usage in your textbook.

Sponsored Links