The \(\mathrm{MnO}_{2}\) used in alkaline batteries can be produced by an electrochemical process of which one half-reaction is $$\mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-}$$ If a current of \(25.0 \mathrm{~A}\) is used, how many hours are needed to produce \(1.00 \mathrm{~kg}\) of \(\mathrm{MnO}_{2} ?\) At which electrode is the \(\mathrm{MnO}_{2}\) formed?

In electrochemical cells, chemical reactions generate electrical energy or vice versa. Here, the production of \(\text{MnO}_{2}\) relies on an electrochemical reaction in an alkaline battery.
The key part of this reaction is a half-reaction, where electrons are either gained or lost:

• Oxidation (loss of electrons) occurs at the anode.
• Reduction (gain of electrons) happens at the cathode.

In this exercise, the half-reaction involved is:
\(\text{Mn}^{2+}(a q)+2 \text{H}_{2} \text{O}(l) \rightarrow \text{MnO}_{2}(s)+4 \text{H}^{+}(a q)+2 \text{e}^{-}\)
This reaction shows electrons being lost (oxidation), forming solid \(\text{MnO}_{2}\). Using this understanding helps to determine the location and the type of reaction occurring.

Calculating molar mass is crucial for converting mass to moles. The molar mass is the sum of the atomic masses of all atoms in a molecule. For \(\text{MnO}_{2}\), we identify:

• Manganese (Mn): Atomic mass = 54.94 g/mol
• Oxygen (O): Atomic mass = 16.00 g/mol \( \times 2 = 32.00 g/mol\)
• Total molar mass = 54.94 g/mol + 32.00 g/mol = 86.94 g/mol

Using this, we convert 1.00 kg (1000 g) of \(\text{MnO}_{2}\) to moles:
\(\text{Moles of } \text{MnO}_{2} = \frac{1000 \text{g}}{86.94 \text{g/mol}} \approx 11.50 \text{mol}\)

This calculation forms the basis for further steps in the problem.

In electrochemical reactions, electric charge transfer is essential. It is quantified by the number of moles of electrons involved. Given the reaction:
\(\text{Mn}^{2+}(a q)+2 \text{H}_{2} \text{O}(l) \rightarrow \text{MnO}_{2}(s)+4 \text{H}^{+}(a q)+2 \text{e}^{-}\), each mole of \(\text{MnO}_{2}\) formation involves 2 moles of electrons. Using the moles of \(\text{MnO}_{2}\) calculated:
\(\text{Total electrons} = 11.50 \text{mol} \( \times 2 \text{mol e}^{-}/\text{mol} = 23.00 \text{mol e}^{-}\)\)
With the Faraday constant (charge of one mole of electrons) being 96,485 C/mol e\(-\):
\(\text{Total charge} = 23.00 \text{mol e}^{-} \times 96,485 \text{C/mol e}^{-} \approx 2,218,155 \text{C}\)
We need this total charge to understand the time required in the process.

Oxidation and reduction reactions are at the heart of electrochemical processes. Oxidation is the loss of electrons, whereas reduction is the gain. They occur concurrently in electrochemical cells. The half-reaction:
\(\text{Mn}^{2+}(a q)+2 \text{H}_{2} \text{O}(l) \rightarrow \text{MnO}_{2}(s)+4 \text{H}^{+}(a q)+2 \text{e}^{-}\)
involves oxidation (loss of electrons), so \(\text{MnO}_{2}\) forms at the anode.
Using the total charge (2,218,155 C) and the current (25.0 A), we calculate the time:
\(t = \frac{\text{Total charge}}{I} = \frac{2,218,155 \text{C}}{25.0 \text{A}} \approx 88,726.20 \text{s}\)
Convert to hours:
\(t \approx \frac{88,726.20 \text{s}}{3600 \text{s/hr}} \approx 24.64 \text{ hours}\)
This shows the time needed for producing 1.00 kg of \(\text{MnO}_{2}\).