Question

Given two half reactions as follows: A2+ → 2 A2+ + 3 e− 4 e− + B → B4− What would you multiply each half-reaction by, to cancel out the electrons?

Answer 1

Answer:

In order to cancel the electrons,. the oxidation half reaction must be multiplied by 4 and the reduction half reaction must be multiplied by 3.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

The half cell reactions for the above reaction follows:

Oxidation half reaction:  $A_{2}^+\rightarrow 2A^{2+}+3e^-$

Reduction half reaction:  $B+4e^-\rightarrow B^{4-}$

Thus, in order to cancel the electrons,. the oxidation half reaction must be multiplied by 4 and the reduction half reaction must be multiplied by 3.

Thus,

4*Oxidation half reaction:  $4A_{2}^+\rightarrow 8A^{2+}+12e^-$

3*Reduction half reaction:  $3B+12e^-\rightarrow 3B^{4-}$

Overall reaction:-

$4A_{2}^++3B\rightarrow 8A^{2+}+3B^{4-}$