Write a balanced ion-electron half equation for the reaction which each undergoes when acting as an oxidising agent?

MnO₄⁻ :
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)

O₂ :
In acidic media : O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)
In alkaline media : O₂ + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)

Cr₂O₇²⁻ :
Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l)

MnO4^- + 8H^+ + 5e^- = Mn^2+ + 4H2O

Method ;
The permanganate ion reduces to Mn^2+
The 4 x oxygen become 4 x water.
Since the 4 x water have 8 hydrogens , the there are 8 x H^+ acids.
Balance the charges using 'e^-' (electrons).
On the LHS we have '1-' & 8+'
On the RHS we have '2+'
We have to add sufficient electrons (e^-) to balance.
-1 + 8 + e^- = 2
7 + e^- = 2
e^- = 2 - 7 = -5
So the 'e^-' becomes '5e^-'. to balance the 'half equation.

Similarly the 'Dichromate'.
Cr2O7^2- + 14H^+ + 6e^- = 2Cr^3+ + 7H2O

Method ;
The dichromate ion reduces to Cr^3+
The 7 x oxygen become 7 x water.
Since the 7 x water have 14 hydrogens , the there are 14 x H^+ acids.
Balance the charges using 'e^-' (electrons).
On the LHS we have '2-' & 14+'
On the RHS we have '2 x 3+ = 6+'
We have to add sufficient electrons (e^-) to balance.
-2 +14 + e^- = 6
12 + e^- = 6
e^- = 6 - 12 = -6
So the 'e^-' becomes '6e^-'. to balance the 'half equation.

Hope that helps!!!!