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 of these
Text Solution
Verified by Experts
The correct Answer is:
C
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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)(g) (1atm)|H^(+), pH = x || Normal calomal electrode has EMF of 0.64 volt at 25^(@)C . The standard reduction potential of normal calomal electrode is 0.28V . What is the pH of solution in anodic compartment. Take (2.303RT)/(F) = 0.06 at 298K .
For the cell Pt_((s)), H_(2) (1 atm)|H^(+) (pH=2)||H^(+) (pH=3)|H_(2) (1atm),Pt The cell reaction is
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 designed as Pt_(H_(2))|HCl_(aq)||Hg_(2)Cl_(2),0.01 N KCl|Hg has emf of 0.271 V at 298 K and 0.2669 at 308 K . The E_(H_(2)^(2+)//Hg) is 0.260 V The heat of reaction for redox change is:
The cell designed as Pt_(H_(2))|HCl_(aq)||Hg_(2)Cl_(2),0.01 N KCl|Hg has emf of 0.271 V at 298 K and 0.2669 at 308 K . The E_(H_(2)^(2+)//Hg) is 0.260 V The E^(@) for oxidation electrode at 298 K is:
Determine at 298 for cell: Pt|Q, QH_(2), H^(+) ||1M KCI |Hg_(2)CI_(2)|Hg(l)|Pt (a) It's emf when pH = 5.0 (b) the pH when E_(cell) = 0 (c) the positive electrode when pH = 7.5 given E_(RP(RHS))^(@) = 0.28, E_(RP(LHS))^(@) = 0.699