Question

In the electrochemical cell using the redox reaction below, the anode half reaction is ________. Sn4+ (aq) + Fe (s) → Sn2+ (aq) + Fe2+ (aq) In the electrochemical cell using the redox reaction below, the anode half reaction is ________. (aq) + (s) (aq) + (aq) Fe→Fe2++2e− Sn4+→Sn2++2e− Fe+2e−→Fe2+ Sn4++2e−→Sn2+ Fe+2e−→Sn2+ Request Answer

Answer 1

Answer:

The anode half reaction is : $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}$

Explanation:

In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.

In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.

We can split the given cell reaction into two half-cell reaction such as-

Oxidation (anode): $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}$

Reduction (cathode): $Sn^{4+}(aq.)+2e^{-}\rightarrow Sn^{2+}(aq.)$

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overall: $Fe(s)+Sn^{4+}(aq.)\rightarrow Fe^{2+}(aq.)+Sn^{2+}(aq.)$

So the anode half reaction is : $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}$