The cell `Pt|H_(2)(g,01` bar) `|H^(+)(aq),pH=x||Cl^(-)(1M)|Hg_(2)Cl_(2)|Hg|Pt` has emf of 0.5755 V at `25^(@)C` the SOP of calomel electrode is `-0.28

Question

The cell `Pt|H_(2)(g,01` bar) `|H^(+)(aq),pH=x||Cl^(-)(1M)|Hg_(2)Cl_(2)|Hg|Pt` has emf of 0.5755 V at `25^(@)C` the SOP of calomel electrode is `-0.28

The cell `Pt|H_(2)(g,01` bar) `|H^(+)(aq),pH=x||Cl^(-)(1M)|Hg_(2)Cl_(2)|Hg|Pt` has emf of 0.5755 V at `25^(@)C` the SOP of calomel electrode is `-0.28V` then pH of the solution will be
A. 11
B. 4.5
C. 5.5
D. none

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
For normal calomel electrode `E_(RP)=E_(RP)^(@)`
`E_(cell)=(0.28-0)-(0.059)/(2)log(([H^(+)]^(2))/(P_(H_(2))))impliespH=5.5`

4 views Answered 2023-02-05 15:29