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$\ce{Na + Hg2Cl2 —> NaCl + Hg}$

I was told to balance this reaction by using the half reaction method. I am confused to as how you would do that since the $\ce{Cl}$ is with both the $\ce{Na}$ and the $\ce{Hg}$.

I did with the inspection method and got $\ce{2Na + Hg2Cl2 —> 2NaCl + 2Hg}$ as my answer.

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    $\begingroup$ Half reaction method? It is like rubbing behind the right ear by the left hand. $\endgroup$
    – Poutnik
    Jan 7 at 19:08
  • $\begingroup$ Agree with @Poutnik. This could be easily done by hit and trial. $\endgroup$ Jan 8 at 9:55

1 Answer 1

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AS $\ce{Hg2Cl2}$ is a covalent compound, you can write those half reactions as:

\begin{align} \ce{2 Na &-> 2 Na+ + 2 e-}\\ \ce{Hg2Cl2 + 2e- &-> 2 Hg + 2 Cl-}\\ \ce{2 Na+ + 2 Cl- &-> 2 NaCl(s)} \end{align}

Therefore

$$\ce{2 Na + Hg2Cl2 -> 2 Hg + 2 NaCl}$$

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