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Chapter 18: Problem 48

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations that are not already balanced. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{H}^{-}(a q)\) b. \(\mathrm{Au}^{3+}(a q)+\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Au}(s)\)

Short Answer

Expert verified
For the given cells: a. The standard cell potential, E掳(Cell), is -2.414 V, which indicates a non-spontaneous reaction under standard conditions. b. The standard cell potential, E掳(Cell), is 0.699 V, which indicates a spontaneous reaction under standard conditions.

Step by step solution

01

a. Reduction and Oxidation half-reactions for H2

To find E掳 for the given cell, we need to identify the two half-reactions: Oxidation half-reaction: \(\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{H}^{+}(a q)+2e^{-}\) Reduction half-reaction: \(2e^{-}+\mathrm{H}^{-}(a q) \longrightarrow \mathrm{H}_{2}(g)\)
02

a. Standard reduction potentials

From Table 18.1, the standard reduction potential for each half-reaction is: E掳(Oxidation) = 0 V E掳(Reduction) = -2.414 V
03

a. Calculation of E掳 of the cell

Now, let's calculate the E掳 of the overall cell reaction: E掳(Cell) = E掳(Reduction) 鈥� E掳(Oxidation) = (-2.414) - (0) = -2.414 V Since E掳(Cell) < 0, the reaction is non-spontaneous under standard conditions.
04

b. Reduction and Oxidation half-reactions for Au3+ and Ag

To find E掳 for the given cell, we need to identify the two half-reactions: Oxidation half-reaction: \(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+e^{-}\) Reduction half-reaction: \(\mathrm{Au}^{3+}(a q)+3e^{-} \longrightarrow \mathrm{Au}(s)\)
05

b. Standard reduction potentials

From Table 18.1, the standard reduction potential for each half-reaction is: E掳(Oxidation) = 0.799 V E掳(Reduction) = 1.498 V
06

b. Calculation of E掳 of the cell

Now, let's calculate the E掳 of the overall cell reaction: E掳(Cell) = E掳(Reduction) 鈥� E掳(Oxidation) = (1.498) - (0.799) = 0.699 V Since E掳(Cell) > 0, the reaction is spontaneous under standard conditions.

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Key Concepts

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

Standard Electrode Potentials
In electrochemistry, standard electrode potential is crucial for understanding how electrochemical cells work. The standard electrode potential, denoted as \( E^{\circ} \), is a measure of the tendency of a chemical species to be reduced, compared to the standard hydrogen electrode, which has an assigned potential of 0 volts.
Standard conditions assume all gases have a pressure of 1 atm, all solutions have a concentration of 1 M, and the temperature is 298 K.
Electrode potentials are typically given for reduction reactions. This means they measure how likely a species is to gain electrons. The higher the \( E^{\circ} \), the more likely the reaction is to occur as written.
  • Standard conditions are important because they provide a common baseline for comparing different reactions.
  • Reduction potentials can be positive or negative, affecting how they influence the overall cell potential.
  • By looking at the table of standard electrode potentials, one can determine the feasibility of a reaction by evaluating the potential of each possible half-reaction.
Redox Reactions
Redox reactions involve the transfer of electrons between two species. The term 'redox' is a combination of reduction and oxidation.
In these reactions, one species gains electrons (reduction), while another loses electrons (oxidation). Each of these occurrences is represented by half-reactions, and combining these allows us to see the full redox reaction.
Understanding which species is oxidized and which is reduced is key.
  • Oxidation involves the loss of electrons and an increase in oxidation state.
  • Reduction involves the gain of electrons and a decrease in oxidation state.
  • Balancing redox reactions requires ensuring that both mass and charge are balanced.

The electron flow in a redox reaction is what allows for the flow of electric current in electrochemical cells.
Knowing half-reactions helps in calculating the cell potential required to determine if a reaction is spontaneous.
Spontaneity of Reactions
To determine if a reaction is spontaneous, the cell potential \( E^{\circ}_{\text{cell}} \) is calculated. In general:
  • If \( E^{\circ} > 0 \), the reaction is spontaneous under standard conditions.
  • If \( E^{\circ} < 0 \), the reaction is non-spontaneous.
Spontaneity has implications for energy change. A spontaneous reaction releases energy when it occurs.
It's essential to note that a non-spontaneous reaction can still occur if energy is supplied.
For example, in the exercise given:
  • For reaction a, \( E^{\circ}_{\text{cell}} = -2.414 \; V \), meaning it is non-spontaneous.
  • For reaction b, \( E^{\circ}_{\text{cell}} = 0.699 \; V \), meaning it is spontaneous.
Understanding spontaneity is vital for predicting whether a reaction will proceed without external energy.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$\mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \quad \mathscr{E}^{\circ}=0.34 \mathrm{V}$$ $$\mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) \quad \mathscr{E}^{\circ}=-1.20 \mathrm{V}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M,\) and the vanadium compartment contains a vanadium electrode and \(\mathrm{V}^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with 0.0800\(M \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) , resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad K=?$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of 500.0 \(\mathrm{mL} \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell}}\) was observed to be 1.98 \(\mathrm{V}\) . The solution was buffered at a pH of 10.00 . a. Calculate\(\mathscr{E}_{\text {cell}}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K\), for the titration reaction. c. Calculate \(\mathscr{E}_{\text {cell}}\) at the halfway point in the titration.

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0\(M \mathrm{M}^{2+} .\) Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C} .\) Assume that the process $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}$$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)

A solution at \(25^{\circ} \mathrm{C}\) contains \(1.0 M \mathrm{Cd}^{2+}, 1.0 M \mathrm{Ag}^{+}, 1.0 \mathrm{M}\) \(\mathrm{Au}^{3+},\) and 1.0 \(\mathrm{M} \mathrm{Ni}^{2+}\) in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.

What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. 1.0 \(M \mathrm{NiBr}_{2}\) solution b. 1.0 \(M \mathrm{AlF}_{3}\) solution c. 1.0 \(M \mathrm{MnI}_{2}\) solution

A solution at \(25^{\circ} \mathrm{C}\) contains 1.0\(M \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-4} M\) \(\mathrm{Ag}^{+}\). Which metal will plate out first as the voltage is gradually increased when this solution is electrolyzed? (Hint: Use the Nernst equation to calculate \(\mathscr{E}\) for each half-reaction.)
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