Let's look at how you would add in the water and hydroxide ions for the $\ce{MO4^{3-}->MO(OH)}$ reaction.

Step 1: Start with the given metal species, with the metal oxidation states included:

$\ce{M^VO4^{3-}->M^{III}O(OH)}$

Step 2: Add the electrons according to the oxidation states. Here the oxidation state on the metal drops by two, so there must be two electrons on the left. Ignore the fact that the overall charges may be unbalanced, that comes next.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH)}$

Step 3: Now we are ready to attack the charge balance. Add the hydroxide ions, which do not couple with the electron exchange, to balance the charges. Counting out the charges we have going into this step we need five hydroxide ions on the right.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH) + 5OH^-}$

Step 4: Finally add water to balance hydrogen; if you rendered the oxidation states and did the math correctly, the oxygen balance falls into place simultaneously. With six hydrogen atoms to balance we use three molecules if water to get the final answer f ok r this reduction:

$\ce{M^VO4^{3-} + 3H2O + 2e^- ->M^{III}O(OH) + 5OH^-}$

Can you now do the remaining reductions?

Let's look at how you would add in the water and hydroxide ions for the $\ce{MO4^{3-}->MO(OH)}$ reaction.

Step 1: Start with the given metal species, with the metal oxidation states included:

$\ce{M^VO4^{3-}->M^{III}O(OH)}$

Step 2: Add the electrons according to the oxidation states. Here the oxidation state on the metal drops by two, so there must be two electrons on the left. Ignore the fact that the overall charges may be unbalanced, that comes next.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH)}$

Step 3: Now we are ready to attack the charge balance. Add the hydroxide ions, which do not couple with the electron exchange, to balance the charges. Counting out the charges we have going into this step we need five hydroxide ions on the right.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH) + 5OH^-}$

Step 4: Finally add water to balance hydrogen; if you rendered the oxidation states and did the math correctly, the oxygen balance falls into place simultaneously. With six hydrogen atoms to balance we use three molecules of water to get the final answer for this reduction:

$\ce{M^VO4^{3-} + 3H2O + 2e^- ->M^{III}O(OH) + 5OH^-}$

Can you now do the remaining reductions?

Let's look at how you would add in the water and hydroxide ions for the $\ce{MO4^{3-}->MO(OH)}$ reaction.

Step 1: Start with the given metal species, with the metal oxidation states included:

$\ce{M^VO4^{3-}->M^{III}O(OH)}$

Step 2: Add the electrons according to the oxidation states. Here the oxidation state on the metal drops by two, so there must be two electrons on the left. Ignore the fact that the overall charges may be unbalanced, that comes next.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH)}$

Step 3: Now we are ready to attack the charge balance. Add the hydroxide ions, which do not couple with the electron exchange, to balance the charges. Counting out the charges we have going into thus step we need five hydroxide ions on the right.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH) + 5OH^-}$

Step 4: Finally add water to balance hydrogen; if you rendered the oxidation states and did the math correctly, the oxygen balance falls into place simultaneously. With six hydrogen atoms to balance we use three molecules if water to get the final answer f ok r this reduction:

$\ce{M^VO4^{3-} + 3H2O + 2e^- ->M^{III}O(OH) + 5OH^-}$

Can you now do the remaining reductions?

Let's look at how you would add in the water and hydroxide ions for the $\ce{MO4^{3-}->MO(OH)}$ reaction.

Step 1: Start with the given metal species, with the metal oxidation states included:

$\ce{M^VO4^{3-}->M^{III}O(OH)}$

Step 2: Add the electrons according to the oxidation states. Here the oxidation state on the metal drops by two, so there must be two electrons on the left. Ignore the fact that the overall charges may be unbalanced, that comes next.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH)}$

Step 3: Now we are ready to attack the charge balance. Add the hydroxide ions, which do not couple with the electron exchange, to balance the charges. Counting out the charges we have going into this step we need five hydroxide ions on the right.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH) + 5OH^-}$

Step 4: Finally add water to balance hydrogen; if you rendered the oxidation states and did the math correctly, the oxygen balance falls into place simultaneously. With six hydrogen atoms to balance we use three molecules if water to get the final answer f ok r this reduction:

$\ce{M^VO4^{3-} + 3H2O + 2e^- ->M^{III}O(OH) + 5OH^-}$

Can you now do the remaining reductions?