Question

Write half reactions for the following equation? : `Mg_((aq))^(+2) + Mn_((s)) -> Mg_((s)) + Mn_((aq))^(+4)`

Answer 1

Writing the Oxidation States for each reactant and product.

`color(blue)(stackrel"+2"overbrace(Mg^(+2))_((aq))) + color(magenta)(stackrel"0"overbrace(Mn )_((s))) -> color(blue)(stackrel"0"overbrace(Mg)_((s))) + color(magenta)(stackrel"+4"overbrace(Mn^(+4)) _((aq))`

`color(white)(wwww`

We see,
`color(blue)(stackrel"+2"overbrace(Mg^(+2))_((aq)) -> stackrel"0"overbrace(Mg)_((s))`

Balancing the charge in this equation,
`Mg^(+2) + 2e^(-) -> Mg`

`color(white)(wwww`

Also,
`color(magenta)(stackrel"0"overbrace(Mn )_((s)) -> stackrel"+4"overbrace(Mn^(+4)) _((aq))`

Balancing the charge in this equation,
`Mn -> Mn^(+4)+ 4e^-`

As, we know, Oxidation is losing of electrons and reduction is gaining of electrons, and oxidation occurs at anode and reduction at cathode,

`"Anode: " Mn -> Mn^(+4)+ 4e^-`

`"Cathode: " Mg^(+2) + 2e^(-) -> Mg`

Answer 2

See below:

Explanation:

Divide the equation into the reduction reaction and the oxidation reaction by inspecting the ox-numbers.

`Mn(s)` goes from ox-state 0 to +4 so it has lost electrons and is therefore oxidized.

`Mn(s) ->Mn^(+4)(aq)+ 4e^(-)`

`Mg^(2+)` turns to `Mg` so it has been reduced as it has gained two electrons and ox state has changed from +2 to 0.

`Mg^(2+)(aq)+2e^(-) ->Mg(s)`
But we must equalize the number of electrons on both sides so the half equation is:

`2Mg^(2+)(aq)+4e^(-) ->2Mg(s)`

So the overall equation is:
`2Mg^(2+)(aq)+Mn(s)+4e^(-) ->2Mg(s)+4e^(-)+Mn^(+4)(aq)`
And then remove the electrons:

`2Mg^(2+)(aq)+Mn(s)->2Mg(s)+Mn^(+4)(aq)`