David Troyer

2022-01-03

What products are obtained in the electrolysis of molten NaI?

vicki331g8

Step 1
Reduction: $N{a}^{+}\left(aq\right)+{e}^{-}⇒Na\left(s\right)$
Determine the chemical equation for the reduction reaction. In the electrolysis of a pure molten salt, the cation is reduced.
Step 2
Oxidation: $2{I}^{-}\left(aq\right)⇒{I}_{2}\left(s\right)+2{e}^{-}$
Determine the chemical equation for the oxidation reaction. In the electrolysis of a pure molten salt, the anion is oxidized.
Step 3
$2N{a}^{+}\left(aq\right)+2e⇒2Na\left(s\right)$
$2{I}^{-}\left(aq\right)⇒{I}_{2}\left(s\right)+2e$
$2N{a}^{+}\left(aq\right)+2{I}^{-}\left(aq\right)⇒2Na\left(s\right)+{I}_{2}\left(s\right)$
Multiply the reduction equation by 2 so that both equations have equal amounts of electrons. Then add the half-reactions to give the overall reaction, and cancel reactants and products that appear on both sides of the reaction $⇒$.
Therefore, Na and ${I}_{2}$ are the product that are obtained.

kalupunangh

Answer is: sodium (Na) and iodine $\left({I}_{2}\right)$.
First ionic bonds in this salt are separeted because of heat: $NaI\left(l\right)⇒N{a}^{+}\left(l\right)+{I}^{-}\left(l\right)$.
Reaction of reduction at cathode(-): $N{a}^{+}\left(l\right)+{e}^{-}⇒\frac{Na\left(I\right)}{×2}$.
$2N{a}^{+}\left(l\right)+2{e}^{-}⇒2Na\left(l\right)$.
Reaction of oxidation at anode(+): $2{I}^{-}\left(l\right)⇒{I}_{2}\left(l\right)+2{e}^{-}$.
The anode is positive and the cathode is negative.

karton

Step 1
In electrolysis, an electric current will pass through an electrolyte, and a redox reaction will take place in the electrolytic cell. The anode is positive and oxidation takes place at the anode. The cathode is negative and reduction takes place at the cathode.
Step 2
The molten Nal, the $N{a}^{+}$, and ${I}^{-}$ ions are present in a liquid state.
the oxidation half-reaction that takes place at the anode is as shown below:
Oxidation: $2{I}^{-}\left(l\right)⇒{I}_{2}\left(g\right)+2{e}^{-}$
The reduction half-reaction that takes place at the cathode is as shown below:
Reduction: $N{a}^{+}\left(l\right)+{e}^{-}⇒N{a}^{+}\left(l\right)$
to get overall reaction, multiply the reaction half-reaction with 2 and add it to oxidation half-reaction as shown below
$\phantom{\rule{0ex}{0ex}}2N{a}^{+}\left(l\right)+2{e}^{-}⇒2Na\left(l\right)\phantom{\rule{0ex}{0ex}}2{I}^{-}\left(l\right)⇒{I}_{2}\left(g\right)+2{e}^{-}\phantom{\rule{0ex}{0ex}}2N{a}^{+}\left(l\right)+2{I}^{-}\left(l\right)⇒{I}_{2}\left(g\right)+2Na\left(l\right)$
Therefore, the products obtained in the electrolysis of molten Nal are Na(l) and ${I}_{2}\left($g)

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