A) 0.00 V
B) + 0.8274 V
C) - 0.8274V
D) can't be said
Correct Answer: C
Solution :
\[\therefore \] \[{{H}_{2}}O+{{e}^{-}}\to \frac{1}{2}\,{{H}_{2}}+O{{H}^{-}}{{E}^{o}}=P\] 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 \[[{{H}^{+}}]\,[O{{H}^{-}}]=\,{{10}^{-14}}\,={{K}_{w}}\] \[\therefore \] \[0=P-\frac{0.0591}{1}\,\log \,{{10}^{-14}}\] \[\therefore \] \[P=-0.8274\,V\]You need to login to perform this action.
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