Question

The following two half-reactions occur in a standard alkaline battery:

Zn(s) + 2OH^-(aq) → ZnO(aq) + H₂O(l) + 2e^-

MnO₂(s) + H₂O(l) + e^- → MnO(OH)(s) + OH^-(aq)

(a)    Describe how the spontaneous flow of electrons occurs in a simple battery (2 pts).

(b)    Identify the cathode and anode of the standard alkaline battery (2 pts).

(c)    Identify the element that is oxidized and the element that is reduced (2 pts).

(d)    Identify the species that acts as the oxidizing agent and the species that acts as the reducing agent (2 pts).

(e)    Write the overall net ionic equation

Answer 1

(a)    Describe how the spontaneous flow of electrons occurs in a simple battery (2 pts).

In a simple battery, there two electrodes, one is the anode that releases electrons and the other is the cathode that catches electrons. Between them, there must an electrolytic medium that allows the electronic flow from one electrode to another. This electronic flow is electricity. In an alkaline battery, potassium hydroxide is used as the electrolyte.

b) Identify the cathode and anode of the standard alkaline battery

In the anode occurs the oxidation, which is the “loss of electrons”. In the cathode occurs the reduction, which is the opposite process of oxidation.

So anode will be

Zn(s) + 2OH^-(aq) → ZnO(aq) + H₂O(l) + 2e^-

We can see that this semireaction produces electrons

And the cathode

MnO₂(s) + H₂O(l) + e^- → MnO(OH)(s) + OH^-(aq)

(c)    Identify the element that is oxidized and the element that is reduced (2 pts).

The element that is oxidized must be in the anode. As the oxidation results in the loss of electrons, this will change the oxidation state of the element to a more positive oxidation state. In our anode, the oxidized element is Zn, which initially has an oxidation state equal to zero

Zn(s) + 2OH^-(aq) → ZnO(aq) + H₂O(l) + 2e^-

To calculate the oxidation state of Zn in ZnO, we said that the charge of the compound must be equal to the sum of the charges of the components multiplied by its subscripts, respectively. So, the charge of Zn will be our unknown X.

ZnO

x + 1(-2) = 0 (compound charge) -----> x - 2 = 0 -----> x = +2

So, Zn atom is going from 0 to +2, and it makes sense because Zn is losing two electrons in this reaction.

The element that is reduced must be in the cathode. As the reduction results in the gain of electrons, this will change the oxidation state of the element to a more negative oxidation state. In this cathode, the reduced element is Mn. Let’s calculate its oxidation state at the beginning and ending of the reaction.

MnO₂(s) + H₂O(l) + e^- → MnO(OH)(s) + OH^-(aq)

MnO₂

x + 2(-2) = 0 -----> x – 4 = 0 -----> x = +4

MnO(OH)

x + 2(-2) + 1(+1) = 0 -----> x -4 +1 = 0 -----> x – 3 = 0 -----> x = +3

So, Mn atom is going from +4 to +3, and it makes sense because Mn is gaining one electron in this reaction.

d)    Identify the species that acts as the oxidizing agent and the species that acts as the reducing agent (2 pts).

A substance that can catch electrons is known as an oxidant agent, in other words, the reduced element is Mn in this case.

A substance that can release electrons is known as a reducing agent, in other words, the oxidized element is Zn in this case.

(e)    Write the overall net ionic equation

To obtain the overall net ionic equation we should add both semi-reactions, but first, we should make the number of given electrons equal to the number of released electrons. In this case, we multiply the semi-reaction of the cathode per two(2), so we have:

Zn(s) + 2OH^-(aq) → ZnO(aq) + H₂O(l) + 2e^-

2MnO₂(s) + 2H₂O(l) + 2e^- → 2MnO(OH)(s) + 2OH^-(aq)

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Zn(s) + 2MnO₂(s) + H₂O(l) → ZnO(aq) + 2MnO(OH)(s)

Note: two molecules of  H₂O as reagent and one as products is equal to have just one as reagente.

Note: two OH^- as reagents and two OH^- as products is equal to have any OH^- in the net reaction. Same with electrons.