A) 0.059 V
B) 0.59 V
C) 0.00 V
D) 0.51 V
Correct Answer: B
Solution :
\[{{E}_{OP}}=E_{OP}^{o}-\frac{0.059}{1}\log \frac{[{{H}^{+}}]}{{{P}_{{{H}_{2}}}}}\] \\[[{{H}^{+}}]={{10}^{-10}};\,\,{{P}_{{{H}_{2}}}}=1\,\,atm\]; \[{{E}_{OP}}=0.59\,\,V\].You need to login to perform this action.
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