Respuesta :
Balanced Half reactions are:
At anode 2[tex]Cl^{-}[/tex] ==> Cl₂+ [tex]2e^{-}[/tex] + H₂O ==> [tex]ClO^{-}[/tex]+ 2[tex]H^{+}[/tex] + [tex]Cl^{-}[/tex]
At Cathode: 2[tex]H^{+}[/tex] + [tex]2e^{-}[/tex] ==> H₂
Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, [tex]H^{+}[/tex], [tex]Mn^{2+}[/tex] and [tex]Cl^{-}[/tex]
Now at Anode reaction will occur as given:
2[tex]Cl^{-}[/tex] ==> Cl₂+ [tex]2e^{-}[/tex] + H₂O ==> [tex]ClO^{-}[/tex]+ 2[tex]H^{+}[/tex] + [tex]Cl^{-}[/tex] (will occur)
At Cathode:
2[tex]H^{+}[/tex] + [tex]2e^{-}[/tex] ==> H₂ (will occur)
At Cathode:
[tex]Mn^{2+}[/tex] + [tex]2e^{-}[/tex]==> Mn (This reaction will not occur)
The deposition of solid Mn will not occur because in aqueous solution, [tex]H^{+}[/tex]will be reduced before [tex]Mn^{2+}[/tex] .
The reduction potentials for [tex]H^{+}[/tex] is zero whereas reduction potential for [tex]Mn^{2+}[/tex] 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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