• # question_answer ${{H}_{2}}{{O}_{2}}\to 2{{H}^{+}}+{{O}_{2}}+2{{e}^{-}}$; $E{}^\circ =-0.68\,V$. This equation represents which of the following behaviour of ${{H}_{2}}{{O}_{2}}$ A) Reducing B) Oxidising C) Acidic D) Catalytic

As ${{H}_{2}}{{O}_{2}}$ is loosing electrons so it is acting as reducing agent.