Question

Here is the discharge reaction for an alkaline battery: zn(s) + 2mno2(s) + h2o(l)\longrightarrow⟶⟶zn(oh)2(s) + mn2o3(s) which species is reduced as the battery is discharged?

Answer 1

The reaction is 
Zn(s) + 2MnO₂(s) + H₂O(l) ⟶ Zn(OH)₂(s) + Mn₂O₃(s)

The reactants are Zn(s) and MnO₂(s) while Zn(OH)₂(s) and Mn₂O₃(s) products.

Let's see the oxidation state of each compound.
sum of the o.s. of the each element of compound = charge of the compound

O.s of O = -2
O.s of OH⁻ = -1

O.s of Zn(s) = 0

O.s of Mn in MnO₂(s) = x
         x + (-2) * 2 = 0
             x            = +4


O.s of Zn in Zn(OH)₂(s) = a
          a + (-1) * 2 = 0
                 a         = +2

O.s of Mn in Mn₂O₃(s) = b
          2*b + (-2) * 3 = b
                         2b   = 6
                           b   = +3
Since the O.s is reduced in Mn from +4 to +3, the reduced species is Mn₂O₃(s).

Answer 2

Answer:

Explanation:

the reaction is  

Zn(s) + 2MnO₂(s) + H₂O(l) ⟶ Zn(OH)₂(s) + Mn₂O₃(s)

This is the discharge reaction, where Zn is undergoing oxidation and Mn is undergoing reduction.

The anode reaction is:

$Zn(s)+2OH^{-}(aq) ---> ZnO(s) + H_{2} O(l) + 2e$

The cathode reaction is:

$MnO_{2}(s)+H_{2}O(l)+2e--->Mn_{2}O_{3} }+2OH^{-}(aq)$

Thus here MnO₂  is undergoing reduction (and the element undergoing reduction is Mn).