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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