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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Ответ:
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 : link
Ответ:
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
Ответ:
D. the atmosphere or feeling created by a literary work.