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Step-by-Step Solution
Step 1: Identify the relevant half-cell reaction
The cell given is Pt(s) | H2(g, 1 bar) | HCl (aq) | AgCl(s) | Ag(s). The overall reaction for the cell can be represented (in simplified form) as:
\frac{1}{2} \,H_{2} + AgCl \,\to\, H^{+} + Ag + Cl^{-}
Step 2: Apply the Nernst Equation for pH = 1 (Case of Sodium)
For the half-cell AgCl(s)|Ag(s)|Cl–, the standard potential is E^{0}_{AgCl|Ag|Cl^-} = 0.22\,\text{V} . The Nernst equation for the cell is:
E_{\text{cell}} = E_{\text{cell}}^{0} - \frac{0.06}{1}\log\bigl([H^{+}][Cl^{-}]\bigr)
Since pH = 1, we have [H^{+}] = 10^{-1}\,\text{M} and, assuming equal concentrations in HCl solution, [Cl^{-}] = 10^{-1}\,\text{M} . Hence:
[H^{+}][Cl^{-}] = 10^{-1} \times 10^{-1} = 10^{-2} \\
\log\bigl(10^{-2}\bigr) = -2
Therefore,
E_{\text{cell}} = 0.22\,\text{V} - 0.06 \times (-2) = 0.22\,\text{V} + 0.12 = 0.34\,\text{V}
Step 3: Determine the incident radiation energy from the sodium case
The work function of Na is given as 2.3\,\text{eV} . Since the stopping potential (the cell voltage) is 0.34\,\text{eV} , the kinetic energy of the photoelectrons is 0.34\,\text{eV} . Hence, the total photon energy of the incident radiation becomes:
\text{Energy of radiation} = \text{Work function} + \text{K.E. of photoelectrons} \\
= 2.3\,\text{eV} + 0.34\,\text{eV} = 2.64\,\text{eV}
Step 4: Apply the same radiation to potassium
With the same incident radiation (energy 2.64\,\text{eV} ), for potassium (K) whose work function = 2.25\,\text{eV} , the kinetic energy of the emitted photoelectrons is:
\text{K.E.} = 2.64\,\text{eV} - 2.25\,\text{eV} = 0.39\,\text{eV}
To stop these electrons, the required stopping potential (cell voltage) must be 0.39\,\text{V} .
Step 5: Find the new pH required to produce the stopping potential of 0.39 V
We again use the Nernst equation for the same cell but now set E_{\text{cell}} = 0.39\,\text{V} :
0.39 = 0.22 - 0.06\,\log\bigl([H^{+}][Cl^{-}]\bigr)
For the HCl solution, we assume [H^{+}] = [Cl^{-}] . Thus:
[H^{+}][Cl^{-}] = [H^{+}]^{2}
So the equation becomes:
0.39 = 0.22 - 0.06\,\log\bigl([H^{+}]^{2}\bigr) \\
0.39 = 0.22 - 0.06 \times 2\,\log\bigl([H^{+}]\bigr) \\
0.39 = 0.22 - 0.12\,\log\bigl([H^{+}]\bigr)
Rearranging,
0.39 - 0.22 = -0.12\,\log\bigl([H^{+}]\bigr) \\
0.17 = -0.12\,\log\bigl([H^{+}]\bigr) \\
\log\bigl([H^{+}]\bigr) = - \frac{0.17}{0.12} \\
= -1.4167 \approx -1.42
Therefore,
[H^{+}] = 10^{-1.42} \implies \text{pH} = 1.42
Step 6: Express the required pH in the asked format
The question requests a nearest-integer form for \text{pH} expressed as \_\_\_\times 10^{-2} . Since 1.42 = 142\times 10^{-2} , the required answer is:
\boxed{142 \times 10^{-2}}
