KMnO4 + HCl → MnCl2 + Cl2 + H2O + KCl
If the reaction occurs in an acidic solution where water and H+ ions are added to the half-reactions to balance the overall reaction, how many electrons are transferred in the balanced reduction half-reaction?
The answer I got is 5*2 which equals 10
K unchanged
Mn goes from +7 to +2 gained five electrons
Cl goes from -1 to zero (but it is a Cl2).
I agree with 10
This one looks ok to me and I get 10 easily, also. I'm still confused by the previous problem.
To determine the number of electrons transferred in the balanced reduction half-reaction, we can examine the oxidation states of the elements involved in the reaction.
In this case, the manganese (Mn) in KMnO4 has an oxidation state of +7, and in MnCl2, it has an oxidation state of +2. This indicates that the Mn undergoes a reduction from +7 to +2.
To balance the reduction half-reaction, we need to account for the change in oxidation states and electron transfer. The half-reaction for the reduction of Mn can be represented as follows:
Mn7+ + x e- → Mn2+
Where "x" represents the number of electrons transferred.
Now, let's compare the oxidation states of Mn in the two compounds:
KMnO4: +7
MnCl2: +2
To balance the charges in the reduction half-reaction, we can multiply Mn2+ by 5 to get:
5 Mn7+ + 5x e- → 5 Mn2+
Therefore, the number of electrons transferred in the balanced reduction half-reaction is 5.
To determine the number of electrons transferred in the reduction half-reaction, you need to first identify the relevant species involved in the reaction. In this case, the reduction half-reaction is the conversion of MnO4- to Mn2+. Here's how you can balance the reduction half-reaction:
1. Write down the unbalanced equation for the reduction half-reaction:
MnO4- → Mn2+
2. Identify the change in oxidation states for manganese (Mn) in this reaction. In MnO4-, the oxidation state of manganese is +7, and in Mn2+, the oxidation state is +2. Therefore, the change in oxidation state is +7 to +2.
3. Determine the number of electrons needed to balance the change in oxidation state. Each electron carries a charge of -1, so to balance the change of +7 to +2, you need to add 5 electrons (7 - 2 = 5).
So, in the balanced reduction half-reaction, 5 electrons are transferred.
Therefore, the correct answer is 5, not 10.