EMF of the following cell is 0.634 volt at 298 K. ltbr. `Pt|H_2)(1 atm) H^(+)(aq) || KCl(1N)|Hg_(2)Cl_(2)(s)|Hg`
Calculate pH of the anode compartment.
[Given: `E_(Cl^(-)|Hg_(2)Cl_(2)|Hg)^(@)` = 0.28 V and `(2.303 Rt)/(F) = 0.059]`

Calculate pH of the half cell : Pt,H_(2)(1 atm)|H_(2)SO_(4)" "E^(c-)=-0.3V

The pH of LHE in the following cell is : Pt, H_(2)(1atm)|H^(o+)(x M)||H^(o+)(0.1M)|H_(2)(0.1atm)Pt E_(cell)=0.295V .

Calomel (Hg_(2)Cl_(2)) on reaction with ammonium hydroxide gives

The e.m.f of the following cell Pt"|" H_(2)(1 " atm.")//H^(+) (aq "||"Ag^(+) (1M) 1Ag (s) " at "25^(@)C is 0.87V. Calculate the pH of the acid solution. E_("Ag^(+)//Ag)^(Theta)=0.80V, E_(2H^(+)//H_(2))=0.0V

For the following electrochemical cell at 298K Pt(s)+H_(2)(g,1bar) |H^(+) (aq,1M)||M^(4+)(aq),M^(2+)(aq)|Pt(s) E_(cell) = 0.092 V when ([M^(2+)(aq)])/([M^(4+)(aq)])=10^(x) Guven, E_(M^(4+)//M^(2+))^(@) = 0.151V, 2.303 (RT)/(F) = 0.059 The value of x is-

The EMF of a galvanic cell Pt|H_(2)(1 atm)|HCl(1M)|Cl_(2)(g)|Pt is 1.29V . Calculate the partial pressure of Cl_(2)(g) . E^(c-)._(Cl_(2)|Cl^(c-))=1.36V .

Hg_2Cl_2 + NH_4OH to

Pt|H_(2)(1 atm )|H^(+)(0.001M)||H^(+)(0.1M)|H_(2)(1atm)|Pt what will be the value of E_(cell) for this cell

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

Calculate EMF of the following half cells : a. Pt, H_(2)(2 atm)|HCl(0.02M)" "E^(c-)=0V b. Pt , Cl_(2)(10 atm)|HCl(0.1 M)" "E^(-c)=1.36V