A) 3.085
B) 4.085
C) 5.085
D) 6.085
Correct Answer: C
Solution :
\[{{H}_{2}}\xrightarrow{{}}2{{H}^{+}}+2{{e}^{-}}\] \[\because \] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}{{\log }_{10}}\frac{{{[{{H}^{+}}]}^{2}}}{[{{H}_{2}}]}\] \[\therefore \] \[0.3=0-\frac{0.0591}{2}{{\log }_{10}}{{[{{H}^{+}}]}^{2}}\] or \[-\log \,[{{H}^{+}}]=\frac{0.3}{0.0591}=5.085\] \[\therefore \] \[pH-\log [{{H}^{+}}]=5.085\]
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