Question

In an electrolytic process nickel sulfide is oxidized in ? two-step reaction: $$ \begin{array}{r} \mathrm{Ni}_{3} \mathrm{~S}_{2}(s) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{NiS}(s)+2 \mathrm{e}^{-} \\ \mathrm{NiS}(s) \longrightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{S}(s)+2 \mathrm{e}^{-} \end{array} $$ What mass of \(\mathrm{Ni}^{2+}\) is produced in solution by passing a current of 67 A for a period of \(11.0 \mathrm{hr}\), assuming the cell is \(90 \%\) efficient?

Short answer

The mass of \(Ni^{2+}\) produced in solution by passing a current of 67 A for a period of 11.0 hours, with 90% cell efficiency, is \(727.76 \ g\).

Step-by-step solution

Calculate the total charge passed through the cell

To determine the total charge passed through the cell, we will use the formula: Charge (Q) = current (I) * time (t) But first, we need to convert the time given in hours to seconds: \(11.0 hours = 11.0 × 3600s = 39,600 s\) Now, we can calculate the total charge: \(Q = 67 A × 39,600 s = 2,658,200 C\)

Determine moles of Ni^2+ produced

In both reactions, 2 electrons are involved in the production of 1 mole of Ni^2+. Using Faraday's constant, which is the charge of 1 mole of electrons (96,485 C/mol), we can determine the moles of Ni^2+ produced. Moles of Ni^2+ (n) = \(\dfrac{Charge}{2 × Faraday's \ constant}\) Since the cell is only 90% efficient, we need to account for the efficiency by multiplying the charge by 0.90: Effective charge = \(0.90 × 2,658,200 C = 2,392,380 C\) Now, calculate the moles of Ni^2+ produced: n(Ni^2+) = \(\frac{2,392,380 C}{2 × 96,485 C/mol} = 12.40 mol\)

Calculate the mass of Ni^2+ produced

To find the mass of Ni^2+ produced, we need to multiply the moles of Ni^2+ by the molar mass of nickel: Molar mass of nickel = 58.69 g/mol Mass of Ni^2+ produced = \(12.40 \ mol × 58.69 \ g/mol = 727.76 \ g\) So, 727.76 grams of Ni^2+ will be produced in solution when we pass a current of 67 A for a period of 11.0 hours, considering the cell is 90% efficient.

Key concepts

Nickel sulfide oxidation

Nickel sulfide oxidation is a chemical process involving the transformation of nickel in its sulfide form to a positively charged nickel ion, specifically the nickel(II) ion, denoted as \(\mathrm{Ni}^{2+}\). This process is part of a larger electrolytic reaction involving electricity to drive the chemical change. In our example, the oxidation process is expressed in two distinct steps.

• In the first reaction, \(\mathrm{Ni}_{3} \mathrm{~S}_{2}(s)\) changes to one \(\mathrm{Ni}^{2+}(aq)\), two \(\mathrm{NiS}(s)\), and releases two electrons.
• In the second step, \(\mathrm{NiS}(s)\) transfers into another \(\mathrm{Ni}^{2+}(aq)\), yellow sulfur \(\mathrm{S}(s)\), and again frees two electrons.

This sequence shows nickel transitioning from compounds to free ions with the help of passing electrons, which are important in the electrolysis process where the electricity flow makes the chemical transformation possible. Understanding these transformations is crucial when determining the outcome of electrolytic processes.

Faraday's constant

Faraday's constant is a fundamental value used in electrochemistry. It describes the total electric charge in one mole of electrons, around 96,485 coulombs per mole (C/mol). This constant is incredibly useful for translating the amount of electric charge passed through an electrolytic cell into the amount of substance processed or produced.

In electrolysis processes, like the oxidation of nickel sulfide, we use Faraday's constant to establish a relation between charge and chemical occurrences. Each reaction step which produces \(\mathrm{Ni}^{2+}\) involves two electrons. Thus, by knowing how much charge has passed, we can calculate how many moles of \(\mathrm{Ni}^{2+}\) ions are formed. It's a vital tool that allows us to convert an abstract concept of charge into tangible amounts of chemicals produced.

Electrochemical efficiency

In electrochemical reactions, efficiency is not limited to the chemical yield; it also considers how effectively current is used. The term 'electrochemical efficiency' often refers to the percentage of electricity that contributes to the desired chemical reaction. In practice, some of the electrical energy may be lost to side reactions or heat.

When discussing a 90% efficient cell, it means that only 90% of the input electrical energy directly contributes to the formation of products. In the example exercise, this efficiency percentage factors into calculations by scaling the total charge to reflect what's effectively used for nickel ion production. Thus, instead of 100% of the charged electrons converting sulfide to \(\mathrm{Ni}^{2+}\), a factor of 0.90 indicates a more realistic production capacity, ensuring accurate results.

Molar mass calculations

Molar mass calculations are a cornerstone of chemistry, allowing us to convert between moles, which relate to atomic and molecular scale quantities, and grams, our more practical unit of mass. Each element or compound has a distinctive molar mass, often found on a periodic table or calculated by summing atomic masses.

For nickel, the molar mass is about \(58.69 \, g/mol\). When we determine the moles of a substance produced in an electrochemical reaction, we then employ this molar mass to compute the actual mass in grams. In our nickel sulfide example, knowing that \(12.40\) moles of \(\mathrm{Ni}^{2+}\) are produced, multiplying by Nickel's molar mass gives us the total product mass. Such conversions allow us to directly relate the chemistry to practical quantities that can be measured and used in applications.