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

The standard reduction potential at 25^@C of the reaction 2H_2O+2e^(-)

1.	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
Answer» Solution :Let the equilibrium constants for the following equations be as follows: `H_2O+H_2O LEFTRIGHTARROW H_3O^+ +OH^(-) K_1` .......(1) `H_3O^+ +e leftrightarrow 1/2H_2+H_2O K_2`......(2) Adding them we get `H_2O+ e leftrightarrow 1/2H_2 +OH^(-) K_3`........(3) We have now to calculate `K_1` Now for EQN(2) , `E_(H_3O^+)^@=0` `therefore K_2=1(because E^@=0.0591/1log K)`......(Eqn 2) For eqn. (2) `E^@=-0.8277` volt (given) `therefore E^@=0.0591/1 LOG K_3=-0.8277` `logK_3=-0.8277/0.0591, K_3=1 times 10^-14` Further as eqn (3) is the sum of eqns. (1) and (2) we have `K_3=K_1.K_2=K_1` `therefore K_1=1 times 10^-14`