Use the half-reaction method to balance the following equation and then identify the oxidizing and reducing agents: Fe(CN)63-(aq) + Re(s) Fe(CN)64-(aq) + ReO4-(aq) [basic]​

Explanation:

The given chemical equation is:

Fe(CN)63-(aq) + Re(s)-> Fe(CN)64-(aq) + ReO4-(aq)

Consider oxidation half reaction and balance it first in acidic conditions:

Add water on the left side to balance the O-atoms:

Add protons on the right side to balance H-atoms:

To balance the charge add electrons:

------------(1)

Reduction half reaction:

Fe(CN)63-(aq) -> Fe(CN)64-(aq)

Add electrons to balance the charge:

---------------(2)

Multiply equation(2) with seven :

------(3)

Add (1) and (3)

Add 8OH- on both sides:

It becomes:

This is the final equation in the basic medium.

Re(s) is oxidised. So it is the reducing agent.

Fe(CN)63- is reduced.It is the oxidising agent.

h let me think