Question

What is the half-reaction that occurs at the cathode during the electrolysis of an aqueous sodium iodide solution? Half-Reaction E° (V) 2H2O(l) + 2e– → H2(g) + 2OH–(aq) –0.83 O2(g) + 4H+(aq) + 4e– → 2H2O(l) +1.23 Na+(aq) + e– → Na(s) –2.71 I2(s) + 2e– → 2I–(aq) +0.53

Answer 1

Explanation:

Reaction equation showing ionization of NaI is as follows.

     $NaI(aq) \rightarrow Na^{+}(aq) + I^{-}(aq)$

Reaction equation showing ionization of water is as follows.

     $H_{2}O(l) \rightarrow H^{+}(aq) + OH^{-}(aq)$

At cathode, only reduction reactions take place. Hence, reactions taking place at cathode are as follows.

     $Na^{+}(aq) + 1e^{-} \rightarrow Na(s)$      $E^{o}$ = -2.71

Standard reduction potential equation for hydrogen ions is as follows.

     $2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)$    $E^{o}$ = 0.00

If a cation has more negative reduction potential then that ion is difficult to reduce as compared to the cation that has lower negative reduction potential.

As $Na^{+}$ has less electronegative reduction potential than $H^{+}$.

Thus, $H^{+}$ will be reduced easily and you will get the product $H_{2}$ at cathode.