• # question_answer Calculate the standard reduction potential for the reaction, ${{H}_{2}}O+{{e}^{-}}\to \,\frac{1}{2}\,{{H}_{2}}+O{{H}^{-}}$, using Nernst equation and the fact that the standard reduction potential for the reaction; ${{H}^{+}}+{{e}^{-}}\to \frac{1}{2}\,{{H}_{2}}$, is by definition equal to 0.00 V at $25{}^\circ C$. A) 0.00 V                          B) + 0.8274 V C) - 0.8274V       D) can't be said

$\therefore$ ${{H}_{2}}O+{{e}^{-}}\to \frac{1}{2}\,{{H}_{2}}+O{{H}^{-}}{{E}^{o}}=P$ volt (say) Also $\frac{1}{2}\,{{H}_{2}}\to {{H}^{+}}+{{e}^{-}}{{E}^{o}}=0.00\,V$ (given) For  ${{H}_{2}}O\to {{H}^{+}}+O{{H}^{-}}\,{{E}^{o}}=P$ Volt $\therefore$ ${{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{1}\,\log \,[{{H}^{+}}]\,[O{{H}^{-}}]$ At equilibrium ${{E}_{cell}}=0$  and      $[{{H}^{+}}]\,[O{{H}^{-}}]=\,{{10}^{-14}}\,={{K}_{w}}$ $\therefore$    $0=P-\frac{0.0591}{1}\,\log \,{{10}^{-14}}$ $\therefore$    $P=-0.8274\,V$