\[2{{H}_{2}}O\xrightarrow{{}}4{{H}^{+}}+{{O}_{2}}-4{{e}^{-}};\] |
\[2{{H}_{2}}O+2{{e}^{-}}\xrightarrow{{}}{{H}_{2}}+2O{{H}^{-}}\] |
Then pH of |
A) cathodic solution will increase
B) anodic solution will decrease
C) both the solutions will remain practically constant
D) both the solutions will increase
Correct Answer: C
Solution :
[c] Due to buffer action the pH will remain practically constant,You need to login to perform this action.
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