Question

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]​

Answer 1

Step-by-step 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:

`Re(s)->ReO_4^-(aq)`

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

`Re(s)+4H_2O->ReO_4^-(aq)\\`

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

`Re(s)+4H_2O->ReO_4^-(aq)+8H^+`

To balance the charge add electrons:

`Re(s)+4H_2O->ReO_4^-(aq)+8H^++7e^-`------------(1)

Reduction half reaction:

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

Add electrons to balance the charge:

`Fe(CN)_6^3^-(aq) + e^- -> Fe(CN)_6^4^-(aq)`---------------(2)

Multiply equation(2) with seven :

`7 Fe(CN)_6^3^-(aq) + 7e^- -> 7Fe(CN)_6^4^-(aq)` ------(3)

Add (1) and (3)

`7 Fe(CN)_6^3^-(aq) +Re(s)+4H_2O -> 7Fe(CN)_6^4^-(aq)+ReO_4^-(aq)+8H^+`

Add 8OH- on both sides:

`7 Fe(CN)_6^3^-(aq) +Re(s)+4H_2O+8OH^- -> 7Fe(CN)_6^4^-(aq)+ReO_4^-(aq)+8H_2O`

It becomes:

`7 Fe(CN)_6^3^-(aq) +Re(s)+8OH^- -> 7Fe(CN)_6^4^-(aq)+ReO_4^-(aq)+4H_2O`

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.