A) \[2\]
B) \[3\]
C) \[5\]
D) \[4\]
Correct Answer: A
Solution :
: For the given reaction, \[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O\] \[E={{E}^{o}}-\frac{2.303RT}{nF}\log \frac{\text{ }\!\![\!\!\text{ Products }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ Reactants }\!\!]\!\!\text{ }}\] \[1.067=1.33-\frac{0.0591}{6}\log \frac{{{[C{{r}^{3+}}]}^{2}}{{[{{H}_{2}}O]}^{7}}}{[C{{r}_{2}}O_{7}^{2-}]{{[{{H}^{+}}]}^{14}}}\] \[1.067=1.33-9.85\times {{10}^{-3}}\log \frac{{{[15\times {{10}^{-3}}]}^{2}}{{[1]}^{7}}}{[4.5\times {{10}^{-3}}]{{[{{H}^{+}}]}^{14}}}\] \[\frac{-0.263}{-9.85\times {{10}^{-3}}}=2\log [15\times {{10}^{-3}}]-\log [4.5\times {{10}^{-3}}]\] \[-14\log [{{H}^{+}}]\] \[26.7=2\times -1.82+2.34+14pH\] \[26.7=-3.64+2.34+14pH\] \[14pH=26.7+3.64-2.34\] \[pH=\frac{28}{14}=2\]You need to login to perform this action.
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