Answer : The overall balanced reaction will be:
[tex]Mn(aq)+2ClO_2(g)\rightarrow Mn^{2+}(aq)+2ClO_2^-(aq)[/tex]
Explanation :
The half-cell reaction will be:
Anodic cell reaction (oxidation) : [tex]Mn(aq)\rightarrow Mn^{2+}(aq)+2e^-[/tex]
Cathodic cell reaction (reduction) : [tex]ClO_2(g)+1e^-\rightarrow ClO_2^-(aq)[/tex]
In order to balance the electrons, we are multiplying cathodic cell reaction by 2 we get:
Cathodic cell reaction : [tex]2ClO_2(g)+2e^-\rightarrow 2ClO_2^-(aq)[/tex]
Now adding both reaction, we get:
Anodic cell reaction : [tex]Mn(aq)\rightarrow Mn^{2+}(aq)+2e^-[/tex]
Cathodic cell reaction : [tex]2ClO_2(g)+2e^-\rightarrow 2ClO_2^-(aq)[/tex]
The overall reaction will be:
[tex]Mn(aq)+2ClO_2(g)\rightarrow Mn^{2+}(aq)+2ClO_2^-(aq)[/tex]
The overall balanced equation occurring in an anode of Mn(s) electrode and a cathode consisting of a Pt(s) electrode is
Mn(aq)+2ClO_2(g) ----> Mn^{2+} (aq)+2ClO_2^- (aq)
Generally, the equation for the cathode and anode reactions are mathematically given as
Mn(aq) ---> Mn^{2+}(aq)+2e^-
ClO_2(g)+1e^-----> ClO_2^- (aq)
Hence in a bid to balance it we have on the cathode side we add 2
2ClO_2(g)+2e^- ---> 2ClO_2^-(aq)
Therefore,the balance eqtation is
Mn(aq)+2ClO_2(g) ----> Mn^{2+} (aq)+2ClO_2^- (aq)
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