Question

Balance the following skeleton reactions and identify the oxidizing and reducing agents:(b) Fe(CN)₆³⁻(aq) + Re(s) →\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{ReO}_{4}^{-}(a q) [basic]

Answer 1

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-   > ReO4-

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

Re-  +4H20-  > ReO4-

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

Re-  +4H20-  > ReO4-+8H+

To balance the charge add electrons:

Re-  +4H20-  > ReO4-+8H+ 7e-------------(1)

Reduction half reaction: Fe(CN)63-(aq) -> Fe(CN)64-(aq)

Add electrons to balance the charge:

Fe(CN)63-(aq) -  +e- > Fe(CN)64- (aq)---------------(2)

Multiply equation(2) with seven :

7{Fe(CN)63-(aq) -  +7e-} > 7{Fe(CN)64- (aq)} ------(3)

Add (1) and (3)

7Fe(CN)63-(aq) -  +7e-} > 7{Fe(CN)64- (aq)}   +   Re-  +4H20-  > ReO4-+8H+ 7e-

Add 8OH- on both sides:7Fe(CN)63-(aq) -  +7e-} > 7{Fe(CN)64- (aq)}   +   Re-  +4H20-   + OH-  > ReO4-+8H+ 7e- +OH-

It becomes:

7Fe(CN)63-(aq)  + rE +8OH-->OH-  >7{Fe(CN)64- +    ReO4-+  4H20

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.

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