A) 96500 C
B) \[2\times 96500\,C\]
C) 9650 C
D) 96.50 C
Correct Answer: C
Solution :
| \[\underset{(+7)}{\mathop{MnO_{4}^{2-}}}\,\underset{(+6)}{\mathop{MnO_{4}^{-}}}\,\] |
| As per the equation, for 1 mole of\[MnO_{4}^{2-}\], 1 F of electricity is required. Thus, for 0.1 mole of\[MnO_{4}^{2-}\], 0.1 F of electricity is required. |
| Since, 1 F = 96500 C |
| \[\therefore \] \[0.1F=0.1\times 96500C\] |
| = 9650 C |
| Hence, 9650 C of electricity is required to completely oxidise \[MnO_{4}^{2-}\]to\[MnO_{4}^{-}\]. |
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