Given the measured cell potential, E_cell is -0.3571 V at 25 degree C in the fol

Question

Given the measured cell potential, E_cell is -0.3571 V at 25 degree C in the following cell, calculate the H^+ concentration Pt (5)|H_2(g, 0.889 atm)|H^+(aq. ? M)Cd^2+ (aq, 1.00 M)|Cd (s) The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, E, are as follows. You have incorrectly identified the anode and the cathode in this cell. According to shorthand cell notation, the anode is written on the left side of the salt bridge. Therefore, the anodic reaction is the oxidation of H_2(g) to H^+(aq), and the cathodic reaction is the reduction of Cd^2+(aq) to Cd(s).

Explanation / Answer

Ecell = -0.3571 V

find H+ given:

E°cell =-0.403 - 0 =- 0.403 V

so

Apply NErnst equation

Ecell = E°cell - 0.0592/n * log(Q)

Q = [H2] / [H+]^2[Cd+2]

[H2] = P/RT = 0.889 atm / (0.082 atm L / mol K) * 298K = 0.03638 M

Q = [H2] / [H+]^2[Cd+2] = (0.03638 )/([H+]^2* 1)

so..

n = 2 electrons

Ecell = E°cell - 0.0592/n * log(Q)

substitute all

-0.3571 = -0.403 - 0.0592/2 * log((0.03638 )/([H+]^2* 1))

(-0.3571 +0.403 )*2/(-0.0592) = log((0.03638 )/([H+]^2* 1))

-1.550=  log((0.03638 )/([H+]^2* 1))

10^-1.550 = (0.03638 )/([H+]^2

[H+]^2 = 0.03638 / (10^-1.550 )

[H+] = sqrt(1.290811) = 1.136 M

[H+] =

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.