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

The anode half reaction is :

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):

Reduction (cathode):

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overall:

So the anode half reaction is :

43.452

step-by-step explanation:

284 times .153 = 43.452