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

Apr 11, 2018

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))

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

Balancing the charge in this equation,
$M {g}^{+ 2} + 2 {e}^{-} \to M g$

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

Balancing the charge in this equation,
$M n \to M {n}^{+ 4} + 4 {e}^{-}$

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

$\text{Anode: } M n \to M {n}^{+ 4} + 4 {e}^{-}$

$\text{Cathode: } M {g}^{+ 2} + 2 {e}^{-} \to M g$

Apr 11, 2018

See below:

#### Explanation:

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

$M n \left(s\right)$ goes from ox-state 0 to +4 so it has lost electrons and is therefore oxidized.

$M n \left(s\right) \to M {n}^{+ 4} \left(a q\right) + 4 {e}^{-}$

$M {g}^{2 +}$ turns to $M g$ so it has been reduced as it has gained two electrons and ox state has changed from +2 to 0.

$M {g}^{2 +} \left(a q\right) + 2 {e}^{-} \to M g \left(s\right)$
But we must equalize the number of electrons on both sides so the half equation is:

$2 M {g}^{2 +} \left(a q\right) + 4 {e}^{-} \to 2 M g \left(s\right)$

So the overall equation is:
$2 M {g}^{2 +} \left(a q\right) + M n \left(s\right) + 4 {e}^{-} \to 2 M g \left(s\right) + 4 {e}^{-} + M {n}^{+ 4} \left(a q\right)$
And then remove the electrons:

$2 M {g}^{2 +} \left(a q\right) + M n \left(s\right) \to 2 M g \left(s\right) + M {n}^{+ 4} \left(a q\right)$