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Balance the redox reaction below using the half-reaction method. Sn2+(aq) + In(s)Sn(s) + In3+(aq) (a) To show your method, write the balanced half reactions below. Use the smallest integer coefficients possible and show electrons as e- . If a box is not needed, leave it blank. (Coefficients of 1 are not needed). Oxidation half-reaction: + + Reduction half-reaction: + + (b) To show your balanced equation, enter an integer in each of the boxes. If the integer is "1," do enter it even though you would normally not show that in the equation. Use the smallest integer coefficients possible. Sn2+(aq) + In(s) Sn(s) + In3+(aq)

Answer :

Answer:

[tex]2In(s)+3Sn^{2+}(aq.)\rightarrow 2In^{3+}(aq.)+3Sn(s)[/tex]

Explanation:

In an oxidation reaction, electrons are being released from a species.

In a reduction reaction, electrons are being consumed by a species.

Here In release electrons and oxidized into [tex]In^{3+}[/tex] . [tex]Sn^{2+}[/tex] consume electrons and reduced to Sn.

[Oxidation half-reaction: [tex]In(s)\rightarrow In^{3+}(aq.)+3e^{-}[/tex]][tex]\times 2[/tex]

[Reduction half-reaction: [tex]Sn^{2+}(aq.)+2e^{-}\rightarrow Sn(s)[/tex]][tex]\times 3[/tex]

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Overall reaction: [tex]2In(s)+3Sn^{2+}(aq.)\rightarrow 2In^{3+}(aq.)+3Sn(s)[/tex]

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