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

The cell Pt|H_(2)(g) (1bar) |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

For the cell, Pt|Cl_2(g,0.4"bar")|Cl^(-)(aq,0.1M)"||"Cl^(-)(aq),0.01M)|Cl_2(g,0.2"bar")|pt

For the cell, Pt|Cl_2(g,0.4"bar")|Cl^(-)(aq,0.1M)"||"Cl^(-)(aq),0.01M)|Cl_2(g,0.2"bar")|pt

For the cell, Pt|Cl_2(g,0.4"bar")|Cl^(-)(aq,0.1M)"||"Cl^(-)(aq),0.01M)|Cl_2(g,0.2"bar")|pt

Determine of pH : The following cell has a potential of 0.55 V at 25^(@)C : Pt(s)|H_(2)(1 bar)|H^(+)(aq. ? M)||Cl^(-)(1 M)|Hg_(2)Cl_(2)(s)|Hg(l) What is the pH of the solution in the anode compartment? Strategy: First, read the shorthand notation to obtain the cell reaction. Then, calculate the half cell potential for the hydrogen electrode from the observed cell potential and the half cell poten-tial for the calomel reference electrode. Finally, apply the Nernst equation to find the pH .

The cell Pt, H_(2)(1atm)|H^(+)(pH=x)|| normal calomel electrode has an EMF of 0.67 V at 25^(@)C Calculate the pH of the solution. The oxidation potential of the calomel electrode on hydrogen scale is -0.28V

The cell Pt, H_(2) (1atm) H^(+) (pH =x)| Normal calomel Electrode has an EMF of 0.67V at 25^(@)C .Calculate the pH of the solution. The oxidation potential of the calomel electrode on hydrogen scale is -0.28V .

The cell Pt, H_(2) (1atm) H^(+) (pH =x)| Normal calomel Electrode has an EMF of 0.67V at 25^(@)C .Calculate the pH of the solution. The oxidation potential of the calomel electrode on hydrogen scale is -0.28V .