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Balancing Redox Reactions

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For the two unbalanced equation below

-Write the unbalanced half-reactions.
-Identify the species oxidized and the species reduced.
-Identify the oxidizing agent and the reducing agent.
-Balance the equation in acid.
-Balance the equation in base.

(a)H2O2(aq)+Ni^2+(aq) --> Ni^3+(aq) + H2O

(b)Cr2o7^2-(aq) +Sn^2+(aq) --> Cr^3+(aq) + Sn^4+(aq)

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(a)H2O2(aq)+Ni^2+(aq)→ Ni^3+(aq) + H2O

Write the unbalanced• half-reactions.
Identify the species oxidized and the species reduced.•
Identify the oxidizing agent and the reducing agent.•

There are two reactions here. The easiest to pick out is the Ni. Ni changes from +2 to +3, meaning it loses 1 electron:

Ni+2  Ni+3 + e-

Because the nickel is losing an electron, it is being oxidized. This means it is the reducing agent as well.

The other reaction involves the oxygen. Normally, hydrogen counts as +1 for determining charge and oxygen counts as 0. In hydrogen peroxide, each oxygen atom is bonded to 1 oxygen and 1 hydrogen (+1 total), so they must both have a -1 charge in hydrogen peroxide. In water, the oxygen is bonded to two hydrogen atoms (+2), so it must have a -2 charge. The difference in charge is 1, so for every water molecule produced, one electron must be added:

H2O2 + 2 e- 2 H2O

Because the peroxide is gaining electrons, it is being reduced. This means it is the oxidizing agent.

Balance the• equation in acid.

The first step with balancing in acid is adding H+ to balance the hydrogen atoms. The nickel reaction has no hydrogen, so it is already balanced:

Ni+2  Ni+3 + e-

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Solution Summary

This solution explains how to write half-reactions and identify the oxidizing and reducing agents. Also, this solution explains how to balance these reactions in both acidic and basic conditions.