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Chapter 20: Problem 87

(a) \(\mathrm{A} \mathrm{Cr}^{3+}(a q)\) solution is electrolyzed, using a current of \(7.60 \mathrm{~A}\). What mass of \(\mathrm{Cr}(s)\) is plated out after \(2.00\) days? (b) What amperage is required to plate out \(0.250 \mathrm{~mol} \mathrm{Cr}\) from a \(\mathrm{Cr}^{3+}\) solution in a period of \(8.00 \mathrm{~h}\) ?

Short Answer

Expert verified
(a) After 2.00 days, 236.44 g of chromium is plated out from the Cr^{3+}(aq) solution. (b) A constant current of 2.517 A is needed to plate out 0.250 mol of chromium in 8.00 hours.

Step by step solution

01

Part (a): Calculate the mass of plated chromium after 2.00 days

1. Convert the time to seconds. As we know that the applied constant current is 7.60 A and the time for which it is applied is 2.00 days, first we need to convert this time into seconds. 2.00 days 脳 24 hours/day 脳 60 minutes/hour 脳 60 seconds/minute = 172800 seconds 2. Calculate the charge that passed through the electrolyte. Next, we will find the total charge that passed through the electrolyte using the formula: charge (Q) = current (I) 脳 time (t) Q = (7.60 A) 脳 (172800 s) = 1313280 Coulombs 3. Calculate the moles of chromium plated. Now, we use Faraday's constant (F = 96485 C/mol) to find the moles of chromium (Cr) plated by dividing the total charge by the charge needed to plate one mole of chromium. Since chromium ions have a charge of +3, we need three moles of electrons to plate one mole of chromium. moles of Cr = (total charge) / (3 脳 Faraday constant) moles of Cr = 1313280 C / (3 脳 96485 C/mol) = 4.5464 mol 4. Calculate the mass of chromium plated. To get the mass of chromium plated out, multiply the moles of chromium by its molar mass (M = 51.996 g/mol): mass of Cr = (moles of Cr) 脳 (molar mass of Cr) mass of Cr = 4.5464 mol 脳 51.996 g/mol = 236.44 g After 2.00 days, 236.44 g of chromium is plated out.
02

Part (b): Find the amperage to plate out 0.250 mol of chromium in 8.00 hours

1. Convert the time to seconds. Since the desired period is 8.00 hours, first, we need to convert this time into seconds. 8.00 hours 脳 60 minutes/hour 脳 60 seconds/minute = 28800 seconds 2. Calculate the charge needed to plate 0.250 mol of chromium. Now, we will find the total charge needed to plate 0.250 mol of chromium by multiplying the number of moles with the charge needed to plate one mole of chromium (3 脳 Faraday constant): charge needed = (0.250 mol) 脳 (3 脳 96485 C/mol) = 72487.5 C 3. Calculate the required current. To find the required current, divide the charge needed by the time available (8.00 h in seconds): current = 72487.5 C/ 28800 s = 2.517 A A constant current of 2.517 A is needed to plate out 0.250 mol of chromium in 8.00 hours.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Laws of Electrolysis
Understanding Faraday's laws of electrolysis is essential when it comes to calculating the quantities involved in the electrolytic deposition of metals. Electrolysis is the process of causing a chemical reaction to occur by passing an electric current through a substance.

Faraday's first law states that the amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (the number of coulombs) that passes through the electrolyte. Mathematically, it can be expressed as \( m = (Q / F) \cdot (M / z) \), where:\
  • \( m \) is the mass of the substance deposited or liberated,
  • \( Q \) is the total electric charge passed through the substance,
  • \( F \) is Faraday's constant (approximately 96485 C/mol),
  • \( M \) is the molar mass of the substance, and
  • \( z \) is the valency number of ions of the substance (the number of electrons needed to reduce or oxidize the ion).
Faraday's second law states that when the same amount of electric current passes through different electrolytes, the mass of substances produced at the electrodes is directly proportional to their equivalent weights.

These laws are pivotal in solving electrolysis calculations like the textbook example, where the mass of chromium plated out after a certain period is determined using the charge passed through the electrolyte, Faraday's constant, and the valency number of chromium.
Molar Mass Calculation
In chemistry, the molar mass calculation is a fundamental concept used to determine the mass of a substance based on its amount in moles. The molar mass is defined as the mass of one mole of a substance, and it is generally expressed in grams per mole (g/mol).

The molar mass can be calculated by summing the atomic masses of all atoms in a molecule. The atomic masses are found on the periodic table and usually take into account the relative abundances of the various isotopes of an element. For instance, the molar mass of chromium (\( Cr \) in this case) is the atomic weight of chromium, which is 51.996 g/mol.

In the electrolysis process, once you find the number of moles of chromium that will be deposited (using Faraday's laws), the next step is to calculate the mass of the chromium by multiplying the number of moles by the molar mass of chromium. This calculation forms a crucial bridge between the abstract concept of moles and the tangible mass of a substance that can be measured or observed.
Stoichiometry
Stoichiometry is the area of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass where the total mass of the reactants equals the total mass of the products. A key concept in stoichiometry is the stoichiometric coefficient, which represents the number of moles of a substance that participate in the reaction.

In the context of electrolysis calculations, stoichiometry helps in understanding the proportion between the electrons transferred and the amount of substance deposited. For example, the chromium ions (\( Cr^{3+} \) in the exercise) require three electrons (\( e^- \) to be reduced to solid chromium (\( Cr(s) \) during electrolysis. This follows the stoichiometric equation: \( Cr^{3+} + 3e^- \rightarrow Cr(s) \).

Thus, to deposit one mole of chromium, three moles of electrons are necessary. Stoichiometry allows the calculation of the necessary moles of electrons, and therefore the required charge, which can then be related back to Faraday's law to find out how long it will take to deposit a certain mass of chromium given a specific current, or vice versa.

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Most popular questions from this chapter

Two wires from a battery are tested with a piece of filter paper moistened with \(\mathrm{NaCl}\) solution containing phenolphthalein, an acid-base indicator that is colorless in acid and pink in base. When the wires touch the paper about an inch apart, the rightmost wire produces a pink coloration on the filter paper and the leftmost produces none. Which wire is connected to the positive terminal of the battery? Explain.

Given the following reduction half-reactions: \(\mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q)\) \(E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V}\) \(\mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\) \(E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V}\) \(\mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V}\) \(\mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) \(E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V}\) (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}{ }^{2-}(a q)\), by \(\mathrm{N}_{2} \mathrm{O}(a q)\), and \(\mathrm{by} \mathrm{VO}_{2}{ }^{+}(a q)\). (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons. (b) For the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q)-\cdots\) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\) is the reducing agent and \(\mathrm{Co}^{2+}(a q)\) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.

The capacity of batteries such as thetypical AA alkaline battery is expressed in units of milliamp-hours (mAh). An "AA" alkaline battery yields a nominal capacity of \(2850 \mathrm{mAh}\). (a) What quantity of interest to the consumer is being expressed by the units of mAh? (b) The starting voltage of a fresh alkaline battery is \(1.55 \mathrm{~V}\). The voltage decreases during discharge and is \(0.80 \mathrm{~V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

(a) Based on standard reduction potentials, would you expect copper metal to oxidize under standard conditions in the presence of oxygen and hydrogen ions? (b) When the Statue of Liberty was refurbished, Teflon spacers were placed between the iron skeleton and the copper metal on the surface of the statue. What role do these spacers play?
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