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

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

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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

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