JEE MAIN - Chemistry (2020 - 3rd September Morning Slot - No. 7)
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
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
142
Explanation
$${1 \over 2}$$H2 + AgCl $$ \to $$ H+ + Ag + Cl-
From Nernst Equation,
Ecell = $$E_{cell}^0$$ - $${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$$
= 0.22 ā 0.06 log 10ā2 = 0.34 V
Work function of Na metal = 2.3 eV
KE of photoelectron = 0.34 eV
Energy of incident radiation = 2.3 + 0.34 = 2.64 eV
For K atom,
Energy of incident radiation for K metal = 2.64 eV
Work function of K metal = 2.25 eV
KE of photoelectrons = 2.64 ā 2.25 = 0.39 eV
$$ \therefore $$ Ecell = 0.39 V
Using Nernst Equation,
0.39 = 0.22 - $${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$$
As $$\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]$$
0.39 = 0.22 - $${{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}$$
$$ \Rightarrow $$ 0.39 = 0.22 - $$0.12 \times \log \left[ {{H^ + }} \right]$$
$$ \Rightarrow $$ 0.39 = 0.22 + 0.12 $$ \times $$ pH
$$ \Rightarrow $$ pH = 1.42 = 142 $$ \times $$ 10-2
From Nernst Equation,
Ecell = $$E_{cell}^0$$ - $${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$$
= 0.22 ā 0.06 log 10ā2 = 0.34 V
Work function of Na metal = 2.3 eV
KE of photoelectron = 0.34 eV
Energy of incident radiation = 2.3 + 0.34 = 2.64 eV
For K atom,
Energy of incident radiation for K metal = 2.64 eV
Work function of K metal = 2.25 eV
KE of photoelectrons = 2.64 ā 2.25 = 0.39 eV
$$ \therefore $$ Ecell = 0.39 V
Using Nernst Equation,
0.39 = 0.22 - $${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$$
As $$\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]$$
0.39 = 0.22 - $${{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}$$
$$ \Rightarrow $$ 0.39 = 0.22 - $$0.12 \times \log \left[ {{H^ + }} \right]$$
$$ \Rightarrow $$ 0.39 = 0.22 + 0.12 $$ \times $$ pH
$$ \Rightarrow $$ pH = 1.42 = 142 $$ \times $$ 10-2
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