The red-ox reaction is MgSO4+2K yields K2 SO4+Mg. I thought the oxidation reaction is 2 k yields K2(subscript) + e^-2. I said the reduction half reaction was Mg+e- yields Mg. Are these correct? Thanks for checking.
close but no cigar.
K ==> K^+ + e or if you want to write it double then
2K ==> 2K^+ + 2electrons. Note it is 2e and not e^-2 (e^-2 denote that the electron has a charge of -2 but it has a charge of -1). You could write 2e^-1 if you wish.
Mg^+2 + 2e ==> Mg is the reduction reaction. Note also that I usually write redox and don't use a hyphen but I don't know what you've been taught by your teacher. Follow your teacher's instructions on this.
To determine if your proposed half-reactions are correct, let's break down the redox reaction:
MgSO4 + 2K → K2SO4 + Mg
First, let's assign oxidation numbers to each element in the reaction. The oxidation number is a charge assigned to each atom to track the movement of electrons in a chemical reaction.
In this case:
- The oxidation number of K is +1 (since it is an alkali metal).
- The oxidation number of O in SO4 is -2, given that the charge of the SO4 ion is -2. This means that the oxidation number of S must be +6.
- The overall charge of the MgSO4 compound is 0, so the sum of the oxidation numbers of all atoms must be 0. Since we know the oxidation numbers of O and S, the oxidation number of Mg can be calculated. In this case, it is +2.
With this information, let's analyze your proposed half-reactions:
Oxidation: 2K → K2 + 2e^-
You correctly identified the oxidation half-reaction for potassium. The oxidation number of K goes from 0 to +1 as it loses one electron.
Reduction: Mg + e^- → Mg
You also correctly identified the reduction half-reaction for magnesium. The oxidation number of Mg goes from +2 to 0 as it gains two electrons.
Therefore, your proposed half-reactions are indeed correct in this context!
Keep in mind that the overall balanced redox reaction should satisfy the conservation of mass and charge and involve the same number of electrons in both half-reactions.