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3 years ago

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Assume a reaction takes place in a basic solution to form the given products: mno4-(aq) + cl-(aq) --------> mno2(s) + cl2(g)

(unbalanced)
i. Balance the given half reactions for atoms and charge

1 answer:

VARVARA [1.3K]3 years ago

3 0

<span>To balance, we write it in its half reactions and balance the half reactions by adding H2O, OH-, H+ and electrons. We do as follows:

MnO4-(aq) + Cl-(aq) --------> MnO2(s) + Cl2(g)

</span>2 (2H2O + MnO2 --> MnO4- + 4H+ 3e- )
<span>3 (2e- + Cl2 --> 2Cl- )
</span>
4H2O + 2MnO2 -----> 2MnO4- + 8H+ + 6e-
<span>6e- + 3Cl2 -------> 6Cl-
</span>----------------------------------------------------------------------
<span>4H2O + 2MnO2 + 3Cl2 --------> 2MnO4- + 6Cl- + 8H+

Since it is in a basic solution,

</span>8OH- + 2MnO2 + 3Cl2 -----> 2MnO4- + 6Cl- + 4H2O

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$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}$ .

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