The photoelectric current from Na (work function, w 0 = 2 . 3 eV ) is stopped by the output voltage of the cell Pt ( s ) ‖ H 2 ( g , 1 bar ) | HCl ( aq · , pH = 1 ) | AgCl ( s ) ∣ Ag ( s ) the p H of

Question

The photoelectric current from Na (work function, $w_{0} = 2.3 eV$ ) is stopped by the output voltage of the cell

$Pt (s) ‖ H_{2} (g, 1 bar) | HCl (aq \cdot, pH = 1) | AgCl (s) ∣ Ag (s)$
the $p H$ of aq. $HCl$ required to stop the photoelectric current from $K (w_{0} = 2.25 eV)$ , all other conditions remaining the same, is $\dots \dots \dots . \times 10^{- 2}$ (to the nearest integer).

Given $2.303 \frac{RT}{F} = 0.06 V; E_{AgCl / Ag / C l}^{0} = 0.22 V$

TestHub · Accepted Answer

Sodium metal : E = E 0 + KE max ; E call 0 = 0 . 22 V Cell reaction Cathode : AgCl s + e - → Ag s + Cl - aq Anode : 1 2 H 2 g → H + aq + e - Overall ; AgCl s + 1 2 H 2 g → Ag s + H + aq + Cl - aq E cell = E cell 0 - 0 . 06 1 log H + Cl - E cell = 0 . 22 - 0 . 06 1 log 10 - 1 10 - 1 = 0 . 22 + 0 . 12 = 0 . 34 V KE max = E cell = 0 . 34 eV So E = 2 . 3 + 0 . 34 = 2 . 64 eV = Energy of photon incident For potassium metal : E = E 0 + KE max 2 . 64 = 2 . 25 + KE max KE max = 0 . 39 = E cell Cell reaction Cathode : AgCl ( s ) + e - → Ag s + Cl - aq Anode : 1 2 H 2 g → H + aq + e - Overall : AgCl s + 1 2 H 2 g → Ag s + H + aq + Cl - aq E cell = E cell 0 - 0 . 06 1 log H + Cl - 0 . 39 = 0 . 22 - 0 . 12 log H + 0 . 17 = 0 . 12 × pH pH = 17 / 12 = 1 . 4166 = 1 . 42

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