Question

For the reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}(\mathrm{g}),\) describe the voltaic cell in which it occurs, label the anode and cathode,use a table of standard electrode potentials to evaluate \(E_{\text {cell }}^{\circ},\) and balance the equation for the cell reaction.

Short answer

The anode is \(Zn(s)\) and the cathode is \(H^{+}(aq)\). The balanced cell reaction is Zn(s) + 2H^{+}(aq) + 2NO3^{-}(aq) -> Zn^{2+}(aq) + 2H2O(l) + 2NO(g). The standard cell potential \(E_{cell}^{\circ}\) is 0.76 V.

Step-by-step solution

Identify Anode and Cathode

In a voltaic cell, the anode is the electrode where oxidation occurs, and the cathode is where reduction occurs. Zinc (\(Zn\)) is oxidized to \(Zn^{2+}\) (losing 2 electrons), so it is the anode. \(H^{+}\) ions are reduced to \(H_2O\), so the cathode consists of hydrogen ions.

Balance the equation

Balancing the equation:Zn(s) + 2H^{+}(aq) + 2NO3^{-}(aq) -> Zn^{2+}(aq) + 2H2O(l) + 2NO(g)

Look up electrode potentials and calculate \(E_{cell}^{\circ}\)

Using a standard table of electrode potentials:For the oxidation half-reaction (at the anode, Zn -> Zn^{2+} + 2e-), the standard electrode potential is -0.76 V.For the reduction half-reaction (at the cathode, 2H^{+} + 2e- -> H2), the standard electrode potential is 0.00 V.Since the cell potential \(E_{cell}^{\circ}\) is the sum of the cathode potential and the anode potential, \(E_{cell}^{\circ}\) = \(E_{cathode}^{\circ}\) - \(E_{anode}^{\circ}\), that is 0.00 - (-0.76) V = 0.76 V

Key concepts

Standard Electrode Potentials

Understanding standard electrode potentials is crucial for analyzing electrochemical reactions, as they provide the voltage associated with a half-cell under standard conditions. Under these conditions, the concentrations of all aqueous solutions are at 1M, the pressure of any gases is at 1 atm, and the temperature is 298 K. These potentials are measured in volts (V) and indicate how likely a substance is to gain electrons, known as reduction (a positive value), or lose electrons, known as oxidation (a negative value).

For example, in the zinc and hydrogen reaction, the standard electrode potential for zinc is -0.76 V, which means zinc readily loses electrons, suggesting it serves as the anode. The standard potential of hydrogen is 0.00 V, indicative of its function as a cathode in this cell. By comparing these values, one can predict the direction of electron flow and the cell's viability to do electrical work.

Cell Potential Calculation

Calculating the cell potential, often denoted as \(E_{\text{cell}}^{\circ}\), is a key step in determining the efficiency and electrical output of a voltaic cell. It is the difference between the potential of the cathode and the anode, which is calculated using the standard electrode potentials of the respective half-reactions. The formula is \(E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}\).

In this zinc-hydrogen reaction, the cell potential is computed as 0.00 V (potential of hydrogen) minus -0.76 V (potential of zinc), which results in a positive cell potential of +0.76 V. This positive value indicates that the reaction is spontaneous and can generate an electrical current.

Redox Reactions

Redox reactions, shorthand for 'reduction-oxidation' reactions, are chemical processes in which electrons are transferred between two substances. These involve oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. In a voltaic cell, the anode undergoes oxidation and the cathode experiences reduction.

In our zinc-hydrogen cell example, the oxidation occurs when zinc \(Zn\) becomes \(Zn^{2+}\), losing 2 electrons in the process. Simultaneously, reduction occurs as the hydrogen ions \(H^{+}\) gain electrons to form water \(H_{2}O\). Together, these half-reactions make up the complete redox process of the cell.

Anode and Cathode Identification

Identifying the anode and cathode in an electrochemical cell is essential to understanding how the cell operates. The anode is where oxidation occurs, meaning it's the site of electron loss. Conversely, the cathode is where reduction takes place, indicating the gain of electrons.

For the given exercise, zinc serves as the anode because it loses electrons during the reaction, represented as \(Zn \rightarrow Zn^{2+} + 2e^-\). On the other hand, the hydrogen ions at the cathode are reduced, which can be seen in the half-reaction \(2H^{+} + 2e^- \rightarrow H_{2}O\). By identifying these electrodes, one can determine the direction of electron flow which is from anode to cathode.

Chemical Equation Balancing

Balancing chemical equations is vital to accurately reflect the law of conservation of mass, ensuring that the same number of each type of atom appears on both sides of the equation. This is especially important in redox reactions, as the transfer of electrons should be equal in both oxidation and reduction half-reactions.

In the zinc and hydrogen reaction, the equation is balanced by ensuring that the charge and atoms are equal on both sides. Starting with one atom of zinc and two nitrate ions, it's crucial to balance the number of hydrogen atoms and water molecules. You end up with a balanced redox equation: \(Zn(s) + 2H^{+}(aq) + 2NO3^{-}(aq) \rightarrow Zn^{2+}(aq) + 2H2O(l) + 2NO(g)\), thus preserving the mass and charge throughout the reaction.