The photoelectric current from Na (Work function, w0
= 2.3 eV) is stopped by the output voltage of
the cell
Pt(s) | H2
(g, 1 Bar) | HCl (aq., pH =1) | AgCl(s) | Ag(s).
The pH of aq. HCl required to stop the photoelectric current form K(w0
= 2.25 eV), all other
conditions remaining the same, is _______ $\times$ 10-2 (to the nearest integer).
Given, 2.303${{RT} \over F}$ = 0.06 V;
$E_{AgCl|Ag|C{l^ - }}^0$ = 0.22 V
Answer (integer)
142
Solution
${1 \over 2}$H<sub>2</sub> + AgCl $\to$ H<sup>+</sup> + Ag + Cl<sup>-</sup>
<br><br>From Nernst Equation,
<br><br>E<sub>cell</sub> = $E_{cell}^0$ - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
<br><br>= 0.22 – 0.06 log 10<sup>–2</sup> = 0.34 V
<br><br>Work function of Na metal = 2.3 eV
<br><br>KE of photoelectron = 0.34 eV
<br><br>Energy of incident radiation = 2.3 + 0.34 = 2.64 eV
<br><br>For K atom,
<br><br>Energy of incident radiation for K metal
= 2.64 eV
<br><br>Work function of K metal = 2.25 eV
<br><br>KE of photoelectrons = 2.64 – 2.25 = 0.39 eV
<br><br>$\therefore$ E<sub>cell</sub> = 0.39 V
<br><br>Using Nernst Equation,
<br><br>0.39 = 0.22 - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
<br><br>As $\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]$
<br><br>0.39 = 0.22 - ${{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}$
<br><br>$\Rightarrow$ 0.39 = 0.22 - $0.12 \times \log \left[ {{H^ + }} \right]$
<br><br>$\Rightarrow$ 0.39 = 0.22 + 0.12 $\times$ pH
<br><br>$\Rightarrow$ pH = 1.42 = 142 $\times$ 10<sup>-2</sup>
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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