Nonstandard half-cell potentials

Electrolysis of Water and Nonstandard Half-Cell Potentials Before we can analyze the electrolysis products of aqueous salt solutions, we must examine the electrolysis of water itself. Extremely pure water is difficult to electrolyze because very few ions are present to conduct a current. If we add a small amount of a salt that cannot be electrolyzed in water (such as Na2S04), however, electrolysis proceeds rapidly. A glass electrolytic cell with separated gas compartments is used to keep the H2 and O2 gases from mixing (Figure 21.25). At the anode, water is oxidized as the O.N. of O changes from —2 to 0 ... 

Electrochemical cell potential for two nonstandard half-cells... 

QB For this cell because the electrodes are identical, the standard electrode potentials are numerically equal and subtracting one from the other leads to the value c°dl = 0.000 V. However, because the ion concentrations differ, there is a potential difference between the two half cells (non-zero nonstandard voltage for the cell). [Pb2+] = 0.100 M in the cathode compartment. The anode compartment contains a saturated solution of Pbl2. We use the Nemst equation (with n = 2) to determine [Pb2+] in the saturated solution. 

Potential difference, Ecell, between oxidation and reduction half-cells under nonstandard conditions. 

Cell potential Potential difference, (,ell between reduction and oxidation half-cells may be at nonstandard conditions. 

The Nernst equation is also used to calculate the electrode potential for a given half-cell at nonstandard conditions. For example, for the half-cell Fe " " - - e Fe " " which has an = 0.77 V and n= the Nemst equation would be ... 

This problem defines nonstandard conditions that must be addressed using the Nernst equation. Virtually anytime you are given concentrations of electrolytes present in a cell (other than 1 M), you need this equation. This problem also presents the challenge of identifying the reactions involved. Iron will be the anode, but we will need to scan the table of standard reduction potentials to identify a possible cathode reaction. The most likely suspect is the reduction of to H2. Once we know both half-reactions, we can calculate the standard cell potential to fill in the appropriate values in the Nernst equation. 

This is the Nernst equation, after the physical chemist W. Nernst, who derived it at the end of the nineteenth century. As above, n is the number of electrons transferred in the cell reaction (2 in reaction 12.7), 5 the Faraday of charge, R the gas constant, and T the temperature (in kelvins). The constant 2.302 59 is used to convert from namral to base 10 logs. At 25 the quantity 2.30259 RT/3 has the value 0.05916, which is called the Nernst slope. The importance of (12.14) is that it allows calculation of the potentials of cells having nonstandard state concentrations (i.e., real cells) from tabulated values of standard half-cell values or tabulated standard Gibbs energies. 

The seawater will be involved in the oxidation half-cell reduction (i.e., it acts as the reductant) with Eox -(0.600 V)= -0.600 V. When the seawater is in equilibrium with the iron system Eceii = rcd + ox = 0 or red = 0.600 V. For nonstandard concentrations at 298K, the total cell potential developed by the iron system when paired with the hydrogen half-cell is given by Eq. (6.26). However, the hydrogen half-cell generates zero electrode potential. Therefore, the electrode potential developed by the iron system is, from Eq. (6.26)... 

Finding for a Concentration Cell Suppose a voltaic cell has the Cu/Cu2+ halfreaction in both compartments. The cell reaction is the sum of identical half-reactions, written in opposite directions. The standard cell potential, , is zero because the standard electrode potentials are both based on 1 M Cu ", so they cancel. In a concentration cell, however, the concentrations are different. Thus, even though is still zero, the nonstandard cell potential, Sceii depends on the ratio of concentrations, so it is not zero. 

Since cell potential depends not only on the half-reactions occurring in the cell, but also on the concentrations of the reactants and products in those half-reactions, we can construct a voltaic cell in which both half-reactions are the same, but in which a difference in concentration drives the current flow. For example, consider the electrochemical cell shown in Figure 18.12 , in which copper is oxidized at the anode and copper ions are reduced at the cathode. The seeond part of Figure 18.12 depicts this cell under nonstandard conditions, with [Cu ] = 2.0 M in one half-cell and [Cu ] = 0.010 M in the other ... 

In Chapter 8 (pages 119-123) we saw that the position of an equilibrium reaction is affected by changes in concentration, temperature and pressure. Redox equilibria are no different. When we compare the voltage of a standard half-cell, X, with a standard hydrogen electrode, we are measuring E for the half-cell X. If we change the concentration or temperature of half-cell X, the electrode potential also changes. Under these nonstandard conditions we use the symbol E for the electrode potential. 

This is a quantitative problem, so we follow the standard strategy. The problem asks about an actual potential under nonstandard conditions. Before we determine the potential, we must visualize the electrochemical cell and determine the balanced chemical reaction. The half-reactions are given in the problem. To obtain the balanced equation, reverse the direction of the reduction half-reaction with the... 

By convention, the potentials of all half-reactions, E°, are found tabulated for the reduction process under standard conditions of temperature (298.15 K), pressure (1 atm), and solute concentrations (1 molar). For nonstandard conditions, the reduction potentials, and hence the cell voltage, will differ. The concentration dependence on the cell voltage is given by the Nemst equation ... 

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