A) 0.177V
B) 0.3V
C) 0.052V
D) 0.104V
Correct Answer: A
Solution :
\[[{{H}^{+}}]\]of first solution \[={{10}^{-pH}}={{10}^{-3}}\] \[[{{H}^{+}}]\]of second solution\[={{10}^{-6}}\] We know, \[E_{cell}^{o}=-0.059\log \frac{{{10}^{-6}}}{{{10}^{-3}}}\] \[=-0.059\log {{10}^{-3}}\] \[=-0.059\times -3\log 10\] \[=-0.059\times -3\times 1\] \[=+0.177V\]You need to login to perform this action.
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