- #1
djh101
- 160
- 5
Just need some clarification as to whether or not I'm doing this right. 1 drop HCl and several drops KMnO4 added to 2mL NaHSO3.
HSO3- → HSO4- + 2e-
MnO42- + 4e- → MnO22-
2HSO3-(aq) + MnO42-(aq) → 2HSO4-(aq) + MnO22-(aq)
A few questions:
-Since NaHSO3 is not an acid, the hydrogen atom would not be removed in the half reaction, correct?
-Are H+ and H2O part of the half reaction or are they only included as a step to help in creating/balancing the equation(s)?
-Should I have done something with the HCl? There were 4 H+ ions on each side of the reaction, which would cancel out and result in no reaction with the HCl.
-If there is no reaction with the HCl, why did I have to add it to the solution?
HSO3- → HSO4- + 2e-
MnO42- + 4e- → MnO22-
2HSO3-(aq) + MnO42-(aq) → 2HSO4-(aq) + MnO22-(aq)
A few questions:
-Since NaHSO3 is not an acid, the hydrogen atom would not be removed in the half reaction, correct?
-Are H+ and H2O part of the half reaction or are they only included as a step to help in creating/balancing the equation(s)?
-Should I have done something with the HCl? There were 4 H+ ions on each side of the reaction, which would cancel out and result in no reaction with the HCl.
-If there is no reaction with the HCl, why did I have to add it to the solution?