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Consider the cell pt | h2 (1 atm); h+ (? m) || hg2cl2(s); cl− (1 m) | hg 2 h+ + 2 e − → h2 e 0 = 0.00 v hg2cl2 + 2 e − → 2 hg + 2 cl− e 0 = 0.268 v if the measured cell potential for the cell is 0.35 volts, what is the ph of the solution? 1. 1.39 2. less than 1.00 3. 4.74 4. 2.77 5. 5.45 020

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Answer:

1.39

Explanation:

[Hg2Cl2]= 1M

[H^+] = ????

E°cell= 0.35V

E= 0.268 V

Therefore E for the reaction must -0.082 V

n= 2 moles of electrons

From Nernst Equation:

E= E°cell- 0.0592/n log [Red]/[Ox]

0.0268= 0.35- 0.0592/2 log 1/[Ox]^2

-0.082= -0.0296 log 1/[Ox]^2

log 1/[Ox]^2= 0.082/0.0296

log 1/[Ox]^2= 2.77

1/[Ox]^2=Antilog (2.77)

[Ox]^2=1.698×10^-3

[Ox] = 0.0412 M

But pH= -log [H^+]= -log(0.0412)= 1.385

batolisis

The pH of the solution is : ( 1 )  1.39

Given data :

E°cell = 0.35V

E = 0.268 V

moles of electrons ( n ) = 2

Hg₂Cl₂ = 1 M

H⁺ =  ?

Determine the pH value of the solution

applying Nernst Equation

E = E°cell - 0.0592 / n  * log ( Red ) / ( Ox )²

    = 0.35 - 0.0592 / 2 * log ( 1 ) / [ Ox ]²

∴ 1 / ( Ox )² = antilog ( 2.77 )

Hence : Ox  = 0.0412 M .

∴ pH = - log ( H⁺ )

     = - log ( 0.0412 ) = 1.385 ≈ 1.39

Hence we can conclude that the pH of the solution is approximately 1.39

Learn more about pH : https://brainly.com/question/23113055

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