Respuesta :
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:
[tex]Re(s)->ReO_4^-(aq)[/tex]
Add water on the left side to balance the O-atoms:
[tex]Re(s)+4H_2O->ReO_4^-(aq)\\[/tex]
Add protons on the right side to balance H-atoms:
[tex]Re(s)+4H_2O->ReO_4^-(aq)+8H^+[/tex]
To balance the charge add electrons:
[tex]Re(s)+4H_2O->ReO_4^-(aq)+8H^++7e^-[/tex]------------(1)
Reduction half reaction:
Fe(CN)63-(aq) -> Fe(CN)64-(aq)
Add electrons to balance the charge:
[tex]Fe(CN)_6^3^-(aq) + e^- -> Fe(CN)_6^4^-(aq)[/tex]---------------(2)
Multiply equation(2) with seven :
[tex]7 Fe(CN)_6^3^-(aq) + 7e^- -> 7Fe(CN)_6^4^-(aq)[/tex] ------(3)
Add (1) and (3)
[tex]7 Fe(CN)_6^3^-(aq) +Re(s)+4H_2O -> 7Fe(CN)_6^4^-(aq)+ReO_4^-(aq)+8H^+[/tex]
Add 8OH- on both sides:
[tex]7 Fe(CN)_6^3^-(aq) +Re(s)+4H_2O+8OH^- -> 7Fe(CN)_6^4^-(aq)+ReO_4^-(aq)+8H_2O[/tex]
It becomes:
[tex]7 Fe(CN)_6^3^-(aq) +Re(s)+8OH^- -> 7Fe(CN)_6^4^-(aq)+ReO_4^-(aq)+4H_2O[/tex]
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.
