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

Learn more about pH : link

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

they are both synthetic elements

explanation:

d

(b) which contains higher levels of vitamin b, oranges or spinach?

explanation:

you can eliminate a, c, and d because they are questions based on opinions.