The standard potential of the reaction
`H_(2)O + e^(-) rightarrow (1/2)H_(2) + OH^(-)` at 298 K by using `k_(w) (H_(2)O) = 10^(-14)`, is:

The standard reduction potential of the reaction H_(2)O+e^(-) rarr 1/2 H_(2) +OH , at 298 K, is

The reaction 2H_(2)O_(2) rarr 2H_(2)O+O_(2) is r=k[H_(2)O_(2)]

The standard reduction potential of the reaction, 2H_(2)O+2e rarr H_(2)+2OH^(-) is -0.8277V at 25^(@)C . Find the value of the equilibrium constant of the reaction 2H_(2)Orarr H_(3)O^(+)+OH^(-) at the same temperature.

The standard reduction potential at 25^@C of the reaction 2H_2O+2e^(-) leftrightarrow H_2 +2OH^(-) is -0.8277 volt. Calculate the equilibrium constant for the reaction 2H_2O=H_3O^+ +OH^(-) at 25^@C

For the rate of the reaction 2H_(2)O_(2)to2H_(2)O+O_(2),r=K[H_(2)O_(2)] it is

For the equilibrium H_(2) O (1) hArr H_(2) O (g) at 1 atm 298 K

For the equilibrium H_(2) O (1) hArr H_(2) O (g) at 1 atm 298 K