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?

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).