Question

how to find electrode potential of hydroxide ion​

Answer 1

Answer:

This is standard reduction potential of O2 + 2H2O + 4e-→4OH- = +0.40 volts and that of Cl2 + 2e- →2Cl- is + 1.36 volts.

Therefore in standard conditions the reduction of Cl2 to Cl- is more feasible than the reduction of water and oxygen to OH-, so the oxidation of OH- to water and oxygen at the anode appears to be more feasible than the oxidation of chloride ions to chlorine gas.

However the conditions are not standard; the concentration of Cl- ions in this solution is very high (much higher than OH-) which shifts the equilibrium Cl2 + 2e- →2Cl- to the left due to Le Chatelier's principle (as the value of the quotient increases, so the equilibrium shifts to the left to decrease the value of the quotient so that it equals the value of the equilibrium constant, thus restoring original conditions) which makes the standard reduction potential of Cl2 much less positive, therefore the standard electrode potential becomes less positive than +0.40 meaning that the reduction of chlorine gas to chlorine is less feasible than the reduction of oxygen and water to hydroxide ions. Thus, the oxidation of Cl- to Cl2 gas at the anode is more feasible (as the Cl- ions have a greater tendancy to lose electrons), so chlorine gas is evolved at the anode.

Explanation:

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