Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table \(17-1\) a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

Short Answer

Expert verified
For the given half-reactions, a) The balanced cell equation is: \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+8 \mathrm{H}^{+} + 6 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Cr}^{3+} + 3 \mathrm{H}_{2} \mathrm{O}_{2}\] and the standard cell potential \(\mathscr{E}^{\circ}_\mathrm{cell} = 2.01 \: \mathrm{V}\). b) The balanced cell equation is: \[3 \mathrm{H}_{2} + 2 \mathrm{Al}^{3+} \rightarrow 6 \mathrm{H}^{+} + 2 \mathrm{Al}\] and the standard cell potential \(\mathscr{E}^{\circ}_\mathrm{cell} = 1.66 \: \mathrm{V}\).

Step by step solution

01

Identify oxidation and reduction half-reactions

From table 17-1, we can see that \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is reduced to \( \mathrm{Cr}^{3+}\) and \(\mathrm{H}_{2} \mathrm{O}_{2}\) is reduced to \(\mathrm{H}_{2} \mathrm{O}\). Since both half-reactions are reductions, we need to reverse one of them to create an oxidation; we can reverse the second half-reaction because it has a smaller reduction potential. So the oxidation half-reaction will be: \[2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\]
02

Balance the number of electrons

The Chromium reduction half-reaction involves 6 electrons, while our new oxidation half-reaction involves 2 electrons. To balance the number of electrons in both half-reactions, we need to multiply the second oxidation half-reaction by 3: \[6 \mathrm{H}_{2} \mathrm{O} \rightarrow 3 \mathrm{H}_{2} \mathrm{O}_{2}+6 \mathrm{H}^{+}+6 \mathrm{e}^{-}\]
03

Write the balanced cell equation

Adding the reduced and oxidized half-reactions: \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} + 6 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O} + 3 \mathrm{H}_{2} \mathrm{O}_{2}+6 \mathrm{H}^{+}+6 \mathrm{e}^{-}\] Simplifying: \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+8 \mathrm{H}^{+} + 6 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Cr}^{3+} + 3 \mathrm{H}_{2} \mathrm{O}_{2}\]
04

Calculate the standard cell potential

Using the reduction potentials from Table 17-1, \(\mathscr{E}^{\circ}_\mathrm{Cr^{3+}}=1.33 \mathrm{V}\) and \(\mathscr{E}^{\circ}_\mathrm{H2O2}=0.68 \mathrm{V}\) , we calculate the standard cell potential: \[ \mathscr{E}^{\circ}_\mathrm{cell} = \mathscr{E}^{\circ}_\mathrm{Cr^{3+}} -(-\mathscr{E}^{\circ}_\mathrm{H2O2}) = 1.33 -(-0.68) = 2.01 \mathrm{V} \] b)
05

Identify oxidation and reduction half-reactions

From table 17-1, we can see that \(\mathrm{H}^{+}\) is reduced to \(\mathrm{H}_{2}\) and \(\mathrm{Al}^{3+}\) is reduced to \(\mathrm{Al}\). Since both half-reactions are reductions, we need to reverse one of them to create an oxidation; we can reverse the first half-reaction (which has a smaller reduction potential) to create an oxidation: \[ \mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \]
06

Balance the number of electrons

The Aluminum reduction half-reaction involves 3 electrons, while our new oxidation half-reaction involves 2 electrons. To balance the number of electrons in both half-reactions, we should multiply the first half-reaction by 3 and the second half-reaction by 2: \[ 3 \mathrm{H}_{2} \rightarrow 6 \mathrm{H}^{+}+6 \mathrm{e}^{-} \] \[ 2 \mathrm{Al}^{3+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al} \]
07

Write the balanced cell equation

Adding the reduced and oxidized half-reactions: \[ 3 \mathrm{H}_{2}+ 2 \mathrm{Al}^{3+}+6 \mathrm{e}^{-} \rightarrow 6 \mathrm{H}^{+}+6 \mathrm{e}^{-} + 2 \mathrm{Al} \] Simplifying: \[ 3 \mathrm{H}_{2} + 2 \mathrm{Al}^{3+} \rightarrow 6 \mathrm{H}^{+} + 2 \mathrm{Al} \]
08

Calculate the standard cell potential

Using the reduction potentials from Table 17-1, \(\mathscr{E}^{\circ}_\mathrm{Al}=-1.66 \mathrm{V}\) and \(\mathscr{E}^{\circ}_\mathrm{H2}=0 \mathrm{V}\) , we calculate the standard cell potential: \[ \mathscr{E}^{\circ}_\mathrm{cell} = - \mathscr{E}^{\circ}_\mathrm{Al} - \mathscr{E}^{\circ}_\mathrm{H2} = -(-1.66) - 0 = 1.66 \mathrm{V} \]

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Key Concepts

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

Standard Cell Potential
Understanding the standard cell potential is crucial for any student studying galvanic cells. It refers to the voltage, or electrical potential difference, of a electrochemical cell under standard conditions, which include 25°C, a 1 molar concentration for each ion participating in the reaction, and a partial pressure of 1 bar for any gases involved.

The standard cell potential, denoted as \( \mathscr{E}^{\circ}_{\text{cell}} \), is calculated by taking the difference between the standard reduction potentials of cathode and anode, also expressed as \( \mathscr{E}^{\circ}_{\text{red,cathode}} - \mathscr{E}^{\circ}_{\text{red,anode}} \). This is because the potential of the galvanic cell is essentially the 'driving force' that pushes electrons from the anode to the cathode. A positive standard cell potential indicates a spontaneous reaction.

Consider the provided example: to determine the standard cell potential for the reaction between \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) and \( \mathrm{H}_{2} \mathrm{O}_{2} \) into \( 2 \mathrm{Cr}^{3+} \) and \( 2 \mathrm{H}_{2} \mathrm{O} \), the potentials from Table 17-1 are used in the calculation, showing a cell potential of 2.01V. This value is positive, meaning that the reaction is predicted to proceed spontaneously under standard conditions.
Half-Reactions
In electrochemistry, half-reactions are at the heart of understanding how galvanic cells function. They show the oxidation or reduction processes separately, detailing what happens at each electrode within the cell. These reactions are fundamental because they help in identifying which species is losing electrons (oxidation) and which is gaining electrons (reduction).

For example, in the exercise, one half-reaction involves chromium (\( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) to \( \mathrm{Cr}^{3+} \)) and the other involves hydrogen peroxide (\( \mathrm{H}_{2} \mathrm{O}_{2} \) to \( \mathrm{H}_{2} \mathrm{O} \)). These half-reactions are vital for constructing the overall balanced equation for the cell's electrochemical process and also for understanding the subsequent behaviour such as electron flow and potential differences.
Electron Balancing
To ensure the law of conservation of charge is satisfied in an electrochemical cell reaction, it is essential to balance the electrons in both the oxidation and reduction half-reactions. Electron balancing is the key step in obtaining a complete equation for the cell reaction.

The procedure involves equalizing the number of electrons lost in the oxidation half-reaction with the number gained in the reduction half-reaction. This is done by multiplying each half-reaction by appropriate coefficients. In the exercise solution, the oxidation half-reaction for hydrogen was multiplied by 3 to match the 6 electrons needed in the reduction of chromium. This practice avoids the difficult task of balancing the entire redox reaction at once, allowing us to focus on balancing electrons in isolation.
Reduction Potentials
The reduction potential, also known as the standard electrode potential, measures the tendency of a chemical species to acquire electrons and be reduced. Each half-reaction has a standard potential listed in an electrochemical series. By convention, all potentials are reported as reductions. To use these potentials for half-reactions that are actually being oxidized in a galvanic cell, one needs to reverse the sign.

In the step-by-step solution, the reduction potentials for \( \mathrm{Cr}^{3+} \) and \( \mathrm{H}_{2} \mathrm{O}_{2} \) are taken from a reference table to calculate the cell potential. If the half-reaction is reversed to reflect oxidation, such as the \( \mathrm{H}_{2} \mathrm{O}_{2} \) to \( \mathrm{H}_{2} \mathrm{O} \) example, reversing the sign of the potential is necessary. This value plays a pivotal role in determining the standard cell potential and, therefore, the feasibility and direction of the electrochemical reaction.

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Most popular questions from this chapter

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned} &3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\ & \mathscr{E}^{\circ}=0.957 \mathrm{V} \end{aligned}$$ $$\begin{aligned} &\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) &\mathscr{E}^{\circ}=0.775 \mathrm{V} \end{aligned}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 \mathrm{M} .\) Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of Al(OH) \(_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} \mathrm{M}\) and the measured cell potential is \(1.82 \mathrm{V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\) $$ \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=? $$

Which of the following statement(s) is/are true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions \(A l^{3+}+3 e^{-} \longrightarrow A l\) and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg},\) aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M\space \mathrm {KF}\) solution b. \(1.0 M\space \mathrm {CuCl}_{2}\) solution c. \(1.0 M \space \mathrm{MgI}_{2}\) solution

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{K}\) is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83} $$ a. Calculate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at \(298 \mathrm{K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

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