In acidic solution, VO_2^+1 is reduced to VO^2+, whereas Mn^2+ is oxidized to MnO_4^-1 .

Given the partial equation
VO_2^+1? +Mn^2+? ---> VO^2+ ? + MnO_4^-1?

What must the coefficients be so that the electrons are balanced?

VO2^+1 ==> VO^2+

Mn^2+ ==> MnO4^-

V changes from +5 on the left to +4 on the right
Mn changes from +2 on the left to +7 on the right.

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To balance the equation and ensure the electrons are balanced, we need to determine the coefficients of the reactants and products.

Here's a step-by-step process to balance the equation and the electrons:

1. Write the unbalanced equation:
VO2+1? + Mn^2+? → VO^2+? + MnO4^-1?

2. Assign variables for the coefficients:
Let's use a, b, c, and d to represent the coefficients for each species in the equation, respectively:
aVO2+1? + bMn^2+? → cVO^2+? + dMnO4^-1?

3. Determine the oxidation states:
The oxidation state for vanadium changes from +5 in VO2+1? to +4 in VO^2+?, and the oxidation state for manganese changes from +2 in Mn^2+? to +7 in MnO4^-1?.

4. Write separate half-reactions for each species:
aVO2+1? → cVO^2+? (half-reaction 1)
bMn^2+? → dMnO4^-1? (half-reaction 2)

5. Balance the elements except oxygen and hydrogen:
In half-reaction 1, we have one vanadium (V) on both sides, so a = c.
In half-reaction 2, we have one manganese (Mn) on both sides, so b = d.

aVO2+1? → aVO^2+? (half-reaction 1)
bMn^2+? → bMnO4^-1? (half-reaction 2)

6. Balance oxygen atoms with water molecules:
In half-reaction 1, the oxygen atoms are already balanced.
In half-reaction 2, the MnO4^-1? ion contains four oxygen atoms. Therefore, we need 4b water molecules on the left side to balance them:
bMn^2+? + 4bH2O → bMnO4^-1? + 8bH+.

7. Balance hydrogen atoms with hydrogen ions (H+):
In half-reaction 1, there are no hydrogen atoms.
In half-reaction 2, we have 8b hydrogen atoms on the right side. Therefore, we need 8b hydrogen ions (H+) on the left side to balance them:
bMn^2+? + 4bH2O → bMnO4^-1? + 8bH+.

8. Balance the charge with electrons:
In half-reaction 1, the charge changes from +5 to +4, so we need one electron (e-) on the left side:
aVO2+1? + e- → aVO^2+?.

In half-reaction 2, the charge changes from +2 to +7, so we need five electrons (5e-) on the left side:
bMn^2+? + 4bH2O + 8bH+ + 5e- → bMnO4^-1? + 8bH+.

9. Cancel out any common terms:
The hydrogen ions (H+) and water molecules (H2O) appear on both sides and can be canceled out:
aVO2+1? + e- → aVO^2+?
bMn^2+? + 5e- → bMnO4^-1?

10. Add the two half-reactions together:
aVO2+1? + bMn^2+? + e- → aVO^2+? + bMnO4^-1?

11. Multiply the half-reactions by appropriate coefficients to equalize the number of electrons (e-):
Since we have one electron in the first half-reaction and five electrons in the second half-reaction, we need to multiply the first half-reaction by five to balance the number of electrons:
5aVO2+1? + 5Mn^2+? + 5e- → 5aVO^2+? + 5bMnO4^-1?

12. Final equation:
5VO2+1? + 5Mn^2+? + 5e- → 5VO^2+? + 5MnO4^-1?

Therefore, the coefficients must be:
5VO2+1? + 5Mn^2+? → 5VO^2+? + 5MnO4^-1?.