The following cell has a potential of 0.45 v at 25°c.

pt(s) | h2(1 atm) | h+(? m) || cl–(1 m) | hg2cl2(s) | hg(l)

the standard half-cell potential for the half-reaction hg2cl2(s) + 2 e– → 2 hg(l) + 2 cl–(aq) is 0.28 v. what is the ph in the anode compartment? the following cell has a potential of 0.45 v at 25°c. pt(s) | h2(1 atm) | h+(? m) || cl–(1 m) | hg2cl2(s) | hg(l) the standard half-cell potential for the half-reaction hg2cl2(s) + 2 e– → 2 hg(l) + 2 cl–(aq) is 0.28 v. what is the ph in the anode compartment?

a. 12.3
b. 4.7
c. 2.9
d. 7.6

C. 2.9

Explanation:

By the Nernst equation:

Ecell = E° - (0.0592/n)*logQ

Where E° = Ecathode - Eanode, n is the number of electrons in the redox reaction (n = 2 in this case), and Q is the reaction coefficient.

The electrode of H₂ has E = 0, an at the way the cell as shown, first we saw the anode (H₂|H⁺) and then the cathode (Cl⁻| Hg₂Cl₂), so

E° = 0.28 - 0 = 0.28 V

The global redox reaction is:

H₂(g) + Hg₂Cl₂(s) → 2H⁺(aq) + 2Cl⁻(aq) + 2Hg(s) (2 electrons are being replaced, because there'll be 2 H⁺ and 2 Cl⁻)

Q = ([H⁺]²*[Cl⁻]²)/pH₂

Q = ([H⁺]²*1/1)

Q = [H⁺]²

0.45 = 0.28 - (0.0592/2)*log[H⁺]²

0.45 - 0.28 = -2*0.0592/2*log[H⁺]       (-log[H⁺] = pH)

0.17 = 0.0592*pH

pH = 2.9

the answer is alpha particles hope i you