Question

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

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

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: \[2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\]

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: \[6 \mathrm{H}_{2} \mathrm{O} \rightarrow 3 \mathrm{H}_{2} \mathrm{O}_{2}+6 \mathrm{H}^{+}+6 \mathrm{e}^{-}\]

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: \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+8 \mathrm{H}^{+} + 6 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Cr}^{3+} + 3 \mathrm{H}_{2} \mathrm{O}_{2}\]

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)

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: \[ \mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \]

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: \[ 3 \mathrm{H}_{2} \rightarrow 6 \mathrm{H}^{+}+6 \mathrm{e}^{-} \] \[ 2 \mathrm{Al}^{3+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al} \]

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: \[ 3 \mathrm{H}_{2} + 2 \mathrm{Al}^{3+} \rightarrow 6 \mathrm{H}^{+} + 2 \mathrm{Al} \]

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: \[ \mathscr{E}^{\circ}_\mathrm{cell} = - \mathscr{E}^{\circ}_\mathrm{Al} - \mathscr{E}^{\circ}_\mathrm{H2} = -(-1.66) - 0 = 1.66 \mathrm{V} \]

Key concepts

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.