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**Problem**: Balance the following equation in basic conditions. Phases are optional.CoCl2 + Na2O2 → Co(OH)3 + Cl- + Na+(redox and oxidation reactions)

###### FREE Expert Solution

###### FREE Expert Solution

First, we must split the reaction into two half-reactions. We do this based on their elements.

$CoCl_2 → Co(OH)_3 + Cl^- \qquad and \qquad Na_2O_2 → Na^+$

The next step is to balance elements different than oxygen and hydrogen first.

$CoCl_2 → Co(OH)_3 + 2\ Cl^- \qquad and \qquad Na_2O_2 → 2\ Na^+$

Next, we balance out oxygens by adding waters or $H_2O$.

$CoCl_2 → Co(OH)_3 + 2\ Cl- \qquad and \qquad Na_2O_2 → 2Na^+ $

$+3\ H_2O \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ +2\ H_2O$

Next, we balance out hydrogens by adding H+.

$CoCl_2 → Co(OH)_3 + 2\ Cl- \qquad and \qquad Na_2O_2 → 2Na^+ $

$+3\ H_2O \qquad \qquad \quad+ 3\ H^{+} \qquad \qquad \quad+ 4\ H^+ \qquad +2\ H_2O$

Next, we balance out the overall charge. For the first reaction:

###### Problem Details

Balance the following equation in basic conditions. Phases are optional.

CoCl_{2 }+ Na_{2}O_{2 }→ Co(OH)_{3 }+ Cl^{-} + Na^{+}

(redox and oxidation reactions)

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