Tag: electrolytic cells and electrolysis

At cathode:
$\displaystyle H _2O\, +\, e^-\, \rightarrow\, \overset{\ominus}O H\, +\, \frac{1}{2} H _2\, (one)$
At anode:
$2HCOO^{\ominus}\, \rightarrow\, 2CO _2\, +\, H _2\, (two)$
$H _2\, and\, CO _2$ at anode and $H _2$ at cathode

Upon electrolysis,
At anode:
$2OH^-$ $\rightarrow H _2O + \frac{1}{2}O _2 + 2e^- $
Hence option A is the right answer.

Haber process is used for producing ammonia from nitrogen and hydrogen.
Down's Process. Molten NaCl upon electrolysis will give Na metal and $Cl _2$ gas. Aqueous NaCl will form NaOH,$H _2 , Cl _2$.

If aq. $NaCl$ is electrolyzed using graphite as anode and $Hg$ as cathode then products are $Cl _2$ gas at anode and $Na$ at the cathode.

$Overall\quad reaction :\ { 2H } _{ 2 }O+{ 2Cl }^{ - }+{ 2Na }^{ + }\rightarrow { 2Na }^{ + }+{ 2OH }^{ - }+{ H } _{ 2 }+{ Cl } _{ 2 }$

$NaBr\rightleftharpoons Na^{+} + Br^{-}$
$2H _{2}O + 2e\rightarrow H _{2} + 2OH^{-}$
$Na^{+} + OH^{-} \rightarrow NaOH$ At cathode
$Br^{-}\rightarrow Br + e^{-}$
$Br + Br \rightarrow Br _{2}$ At anode
So the products are $H _{2}$ and $NaOH$ (at cathode) and $Br _{2}$ (at anode).

When water is added, HCl, KOH, NaI and $CaCl _2$ will produce an electrolytic solution. However, $N _2$ will not produce an electrolytic solution on addition of water. Nitrogen is a homo-nuclear diatomic molecule which does not dissociate into aqueous solution.

Electrolysis of aqueous $KCl$ solution will give $KOH$. This is similar to the preparation of $NaOH$ by electrolysis of aqueous $NaCl$ solution. The electrolysis of fused $KOH$ will give potassium metal.

Higher the standard reduction potential, higher will be the tendency to undergo reduction so, more fast will be its deposition on electrode (cathode) .