The photoelectric current from Na (work function, w0 = 2.3 eV) is stopped by the output voltage of t — JEE Mains Chemistry Past Papers Chemistry Question
Question
The photoelectric current from Na (work function, w0 = 2.3 eV) is stopped by the output voltage of the cell Pt(s)|H2(g, 1bar) | HCl(aq., pH = 1) | AgCl(s) | Ag(s) the pH of aq. HCl required to stop the photoelectric current from K(w0 = 2.25 eV), all other conditions remaining the same, is ……… × 10–2 (to the nearest integer). Given F RT . = 0.06 V; V . E0 Cl | Ag | AgCl
💡 Solution & Explanation
Sodium metal : E = E0 + (KE)max ; cell E = 0.22 V Cell reaction Cathode : AgCl(s) + e– Ag(s) + Cl–(aq) Anode : 1 H2(g) H+(aq) + e– _________________________________________________ Overall : AgCl(s) + 1 H2(g) Ag(s) + H+(aq) + Cl– (aq) cell E = cell E – 06 . log [H+] [Cl–] cell E =0.22 – 06 . log [10-1] [10-1] = 0.22 + 0.12 = 0.34 V | JEE MAIN-2020 | DATE : 03-09-2020 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY PAGE # 8 (KE)max = cell E = 0.34 eV So E = 2.3 + 0.34 = 2.64 eV = Energy of photon incident For potassium metal : E = E0 + (KE)max 2.64 = 2.25 + (KE)max (KE)max = 0.39 = Ecell Cell reaction Cathode : AgCl(s) + e– Ag(s) + Cl–(aq) Anode : 1 H2(g) H+(aq) + e– _________________________________________________ Overall : AgCl(s) + 1 H2(g) Ag(s) + H+(aq) +