`Pt |{:((H_(2))),(1atm):}:| pH = 2:|:|:pH =3 |:{:((H_(2))Pt),(1atm):}:|`. The cell reaction for the given cell is:-

For the cell Pt_((s)), H_(2) (1 atm)|H^(+) (pH=2)||H^(+) (pH=3)|H_(2) (1atm),Pt The cell reaction is

Consider the cell |{:(H_(2)(Pt)),(1atm):} :|: {:(H_(3)O^(+)(aq)),(pH =5.5):}:||{:(Ag^(+)),(xM):}:|Ag . The measured EMF of the cell is 1.023 V . What is the value of x ? E_(Ag^(+),Ag)^(@) +0.799 V.[T = 25^(@)C]

Consider the cell : Pt|H_(2)(p_(1)atm)|H^(o+)(x_(1)M) || H^(o+)(x_(2)M)|H_(2)(p_(2)atm)Pt . The cell reaction be spontaneous if

Pt(H_(2))(1atm)|H_(2)O , electrode potential at 298K is

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

The cell Pt(H_(2))(1atm) |H^(+) (pH =?) T^(-) (a=1)AgI(s), Ag has emf, E_(298KK) =0 . The electrode potential for the reaction AgI +e^(-) rarr Ag + I^(Theta) is -0.151 volt. Calculate the pH value:-

An electrochemical cell is shown below Pt, H_(2)(1 "atm")|HCl(0.1 M)|CH_(3)COOH(0.1 M)|H_(2)(1 "atm") , The emf of the cell will not be zero, because

Pt(H_(2))(p_(1))|H^(o+)(1M)|(H_(2))(p_(2)),Pt cell reaction will be exergonic if

Pt(H_(2))(p_(1))|H^(o+)(1M)|(H_(2))(p_(2)),Pt cell reaction will be exergonic if

The emf of the following cell is observed to be 0.118V at 25^(@)C . [Pt,H_(2)(1atm)|HA(100mL 0.1M||H^(o+)(0.1M)|H_(2)(1atm)|Pt] a. If 30mL of 0.2M NaOH is added to the negative terminal of battery, find the emf of the cell. b. If 50mL of 0.2 M NaOH is added to the negative terminal of battery, find the emf of teh cell.