Following cell has EMF 0.7995 V.
`Pt(s)` I `H_2(1 atm)` I `HNO_3(1M)` II `AgNO_3(1M)` I `Ag (s)`.
if we add enough KCl to the Ag cell so that the final chloride concentration is 1M. Now the measured emf of the cell is 0.222 V. The Ksp of AgCl would be approximately

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.

For a cell Ag(s)|AgNO_(3)(0.01M)||AgNO_(3)(1.0M)|Ag(s) (i). Calculate the e.m.f. of the cell at 25^(@)C (ii). Write the net cell reaction. (iii). Will the cell generate e.m.f when two concentrations become equal?

Calculate the emf of the following cell at 25^(@)C Ag(s)|AgNO_(3) (0.01 M)||AgNO_(3) (0.05 M)|Ag (s)

The EMF of the following cellis 1.05V at 25^(@)C: Pt,H_(2)(g)(1.0 atm)|NaOH(0.1m),NaCl(0.1M)|AgCl(s),Ag(s) a. Write the cell reaction, b. Calculate pK_(w) of water.

Which of the following changes will increase the emf of the cell : Co(s)| CoCl_(2) (M_1) || HCl(M_2) | pt(H_(2), g)

The measured voltage of the cell, Pt(s)|H_2(1.0"atm")|H^+(aq)||Ag^+(1.0M)|Ag(s) is 1.02 V " at " 25^@C . Given E_("cell")^@ is 0.80 V, calculate the pH of the solution .

EMF of the following cell is 0.6 volt. Ag (s) | AgBr ( s) | KBr (0.01 m) |AgNO_(3) ( 0.001M) | Ag( s) K_(sp) of AgBr is expressed as 1 xx 10^(-x) , x is [Take ( 2.303RT)/(F ) = 0.06 V ]

The measured voltage for the reaction with the indicated concentration is 1.50 V. Calculate E^(@) . Cr(s)+3Ag^(+)(aq,0.10M)to3Ag(s)+Cr^(3+)(aq,0.30 M)

At 300 k, /_\ H for the reaction Zn(s)+AgCl(s)toZnCl_2(aq)+2Ag(s) is -218 KJ/mol while the e.m.f of the cell was 1.015V. ((dE)/(dT))_p of the cell is :