Question

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

Answer 1

Answer:

pH ˜ 1.39

Explanation:

solution

given the cell equation as follows:

Pt | H2 (1 atm); H+ (? M) || Hg2Cl2(s); Cl- (1 M) | Hg

half of the reactions are

2H+ + 2 e- --> H2 (E0 = 0.00 V )

Hg2Cl2 + 2 e- --> 2Hg + 2Cl- (E0 = 0.268 V)

E0 cell = E0 cathode -E0 anode

= 0.268 V -0.00V ( the more of positive is cathode while the smaller is anode )

= 0.268V

knowing that  Ecell = E0 cell - 0.0592 V/n) log [oxd/red]

0.35 V = 0.268V -(0.0592V/2) log [H+ ]2 /[Cl- ]

0.35V-0.268V =- 0.0296V*2 log [H+ ]/[1.0]

0.082V = -0.0592V log [H+ ]

so 0.082V /0.0592V = - log [H+ ]

1.385 = - log [H+ ]

1.385 = pH (since pH = - log [H+ ] )

In conclusion, pH ˜ 1.39

Answer 2

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