Question

In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn²⁺ half-cell and an H₂/H⁺ half-cell under the following conditions: [Zn²⁺ ] = 0.021 M [H⁺ ]= 1.3 M partial pressure of H₂ = 0.32 atm. Calculate E_cell at 298 K (enter to 3 decimal places).

Zn²⁺ (aq) + 2e ⁻ $LaTeX: \longrightarrow$ ⟶ Zn(s) E° = − 0.76 V

2H⁺ (aq) + 2e ⁻$LaTeX: \longrightarrow$ ⟶    H2(g) E° = 0.00 V

Answer 1

Solution In electrochemical series in iz above Hydrogen (H). Hence, an will reduce H. (note in electrochemical series upper metal reduce lower element/metal) So, Zn will undergoes oxidation & ( loss of electron) H will undergoes reduction oxidation occur at anode & reduction occur at cathod. given, Zn2+ + 2 é - > Zn(s) E= -0'76 V E= -0.76 V Zn(5) -- Zn2+ + 2€ 2H+ + 22 - H2(3)E° = o v Anode reaction :- Zn (s) – RE -> Zo{t(ay) Cathed reaction : a#+(99) + 2e → H2 (8) overlail beaction:- Ins) + 2H+(49) - > 70 et Hz (3) herein anstient, [In 2+7, PH2.

Q = [zn 2+] x PHz H = partial pressme of H₂ gas, given, [2n++] = 0.021 M TH+] = 1'3 M latm=1013 bar A 0.32 atm ] = 0'32X1'013 L =0.324 bar = 0'32 aton = 0'324 bar cell' Zn/2m++ (0-0211) ** (13M)/Hz (0*32 ctmp), Ecece = Feathed – Eanode. = E p/n - E39/2*** =o-(-0.76) STARTARRATSASSAS = 0.76 V. O'059 Ecele = Ecele - 1959 10g Q, = 0.021 0.324 0'76 - 0.059 log (13) n = number of electron involve in the reaction (halt reaction). = 2 .:. Ecell 0.831 v As N ATIONALE