Question

Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:(b) MnCl₂.

Answer 1

Balanced Half reactions are:

At anode 2$Cl^{-}$  ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$+ 2$H^{+}$ + $Cl^{-}$

At Cathode:  2$H^{+}$ + $2e^{-}$ ==> H₂

Since the question states that you are using an aqueous solution of MnCl₂,  so ions will have present are, H₂O, $H^{+}$, $Mn^{2+}$ and $Cl^{-}$

Now at Anode reaction will occur as given:

2$Cl^{-}$  ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$+ 2$H^{+}$ + $Cl^{-}$  (will occur)

At Cathode:

2$H^{+}$ + $2e^{-}$ ==> H₂ (will occur)

At Cathode:

$Mn^{2+}$ +  $2e^{-}$==> Mn (This reaction will not occur)

The deposition of solid Mn will not occur because in aqueous solution, $H^{+}$will be reduced before $Mn^{2+}$ .

The reduction potentials for $H^{+}$ is zero whereas reduction potential for $Mn^{2+}$ is - 1.18V.

The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.

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