Question

A solution containing a metal ion \(\left(\mathrm{M}^{2+}(a q)\right)\) is electrolyzed by a current of 7.8 A. After 15.5 minutes, \(2.39 \mathrm{~g}\) of the metal is plated out. (a) How many coulombs are supplied by the battery? (b) What is the metal? (Assume \(100 \%\) efficiency.)

Short answer

Question: A metal with \(2+\) ions was deposited on a cathode in an electrolysis experiment. The current was 7.8 A, and the electrolysis lasted for 15.5 minutes. The mass of the deposited metal was 2.39 g. Determine (a) the amount of charge supplied by the battery and (b) the identity of the metal, assuming 100% efficiency. Answer: (a) The amount of charge supplied by the battery is 7254 C. (b) The metal involved in the electrolysis is copper (Cu).

Step-by-step solution

Calculate the total charge supplied by the battery

We know that the amount of charge (Q) is given by the formula: \(Q = It\), where \(I\) is the current and \(t\) is the time for which the current flows. Here, the current is given as 7.8 A and the time is 15.5 minutes. First, we need to convert the time to seconds: \(15.5\,\text{min} = 15.5 × 60\,s = 930\,s\) Now, we can calculate the total charge (Q): \(Q = (7.8\,A) × (930\,s) = 7254\,\text{C}\)

Calculate moles of metal ions involved in the electrolysis

According to Faraday's law of electrolysis, the amount of substance deposited (in moles) is proportional to charge flow during electrolysis: \(n = \frac{Q}{zF}\), where \(n\) is the amount of substance (in moles), \(Q\) is the charge flow, \(z\) is the ion charge, and \(F\) is Faraday's constant (\(96485\,\text{C/mol}\)). Since we're looking for a metal with \(2+\) ions, we can deduce that \(z = 2\). Now, we can calculate the moles of metal ions involved in the electrolysis: \(n = \frac{7254\,\text{C}}{2 × 96485\,\text{C/mol}} = 0.0376\,\text{mol}\)

Determine the molar mass of the metal

From the problem, we know that the mass of the deposited metal is \(2.39\,\text{g}\). We can use this mass and the calculated moles to find the molar mass of the metal: \(\text{Molar mass of metal} = \frac{\text{mass}}{\text{moles}} = \frac{2.39\,\text{g}}{0.0376\,\text{mol}} ≈ 63.6\,\frac{\text{g}}{\text{mol}}\)

Identify the metal

The calculated molar mass of the metal is approximately \(63.6\,\frac{\text{g}}{\text{mol}}\). Referring to the periodic table, we can identify the metal as copper (Cu), which has a molar mass of \(63.5\,\frac{\text{g}}{\text{mol}}\). In conclusion, (a) The amount of charge supplied by the battery is \(7254\,\text{C}\). (b) The metal involved in the electrolysis is copper (Cu).

Key concepts

Faraday's Law of Electrolysis

In understanding the process of electrolysis, Faraday's Law of Electrolysis serves as a fundamental principle. This scientific law quantifies the relationship between the quantity of an element released at an electrode and the amount of electric charge passing through the substance.

According to Faraday's first law, the mass of a substance altered at an electrode during electrolysis is directly proportional to the quantity of electricity that passes through the electrolyte. Mathematically, this relationship is expressed as:
\[ m = (Q / zF) \times M \
\] where \( m \) is the mass of the substance in grams, \( Q \) is the total electric charge in coulombs, \( z \) is the valency number (charge) of the ion (e.g. 2+ for a copper ion), \( F \) is Faraday's constant (approximately \(96485\) C/mol), and \( M \) is the molar mass of the substance in g/mol.

The second law states that when the same amount of charge passes through multiple electrolytes, the masses of elements deposited are directly proportional to their equivalent weights. This law highlights the importance of the substance's valency and atomic mass when interpreting electrolysis results.

During electrolysis, it's essential to pay attention to the efficiency of the process, as some of the charge may not contribute to the desired chemical change and instead produce heat or other side reactions.

Calculating Moles in Electrolysis

One of the critical aspects of electrolysis problems involves translating electric charge into moles of substances involved. To calculate the number of moles (\(n\)) of ions that undergo electrolysis, we use the formula derived from Faraday's law:
\[ n = \frac{Q}{zF} \
\] In this equation, \(n\) represents the amount of substance in moles, \(Q\) is the charge in coulombs, \(z\) is the charge number of ions, and \(F\) stands for Faraday’s constant.

For example, if a current of \(I\) amperes passes through a solution for \(t\) seconds, the charge \(Q\) supplied by the battery can be calculated by \(Q = I \times t\). Current is measured in amperes, which are equivalent to coulombs per second. Therefore, the time (\(t\)) must be expressed in seconds for the units to be consistent.

Once you have \(Q\), and you know \(z\) from the ion's charge, you can calculate the moles of atoms or ions deposited at the electrode during electrolysis. This calculation is crucial for determining the amount of a substance that has undergone electrolysis, facilitating further calculations like determining the molar mass.

Determining Molar Mass

Determining molar mass is a common task in chemistry, significant in electrolysis calculations. The molar mass is the mass per mole of a substance and is reported in units of grams per mole (g/mol). As we address an electrolysis problem, the measure depends directly on both the substance's mass deposited on the electrode and the amount of substance in moles that was involved during the process.

We typically find the molar mass through the formula: \[ \text{Molar Mass} = \frac{\text{mass of deposited metal (g)}}{\text{moles of metal ion}} \
\] To deduce the metal's identity, one also needs to approximate or have the precise value of its molar mass by comparing the computed molar mass with values from a periodic table.

With careful measurement of the mass of the deposited substance and accurate calculation of moles involved based on charge and Faraday's constant, molar mass determination becomes a straightforward application of stoichiometry. This information is invaluable, not just for identifying the substance in question but also for determining purity, and understanding reaction proportions in wider chemical processes.