[" The cell Pt "|H_(2)(g,0.1" bar) "|H^(+)(aq),pH=X|Cl^(-)(aq,lM)|Hg_(2)Cl_(2)|Hg/Pt," has e.m.f.of "0.5755Vat],[25^(@)C." The SOP of calomel electrode is - "0.28V," then pH of solution will be "($)]

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

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 (d)None of these

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

Write the cell reaction of the given galvanic cell : Pt|H_(2)"(g, 1atm)"|HCl(aq)|Hg_(2)Cl_(2)|Hg|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 .

Find the pH of the following acid solution Pt"|"H_(2)(1" bar")"|"H^(+) ("aq. acid")"||"H^(+) (1M) "|"H_(2)(g, 1 "bar")"|"Pt The measured e.m.f. of the cell = 0.182 V.

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 Emf is? (a)0.051V (b)-0.051V (c)0.102V (d)0.0255V