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

Step-by-step 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).

Answer 2

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).