Copper half-cell potential

V, then measure the potential of another half-cell, such as the copper halfcell, with that half-cell. The entire potential of this cell can then be assigned to the other (copper) half-cell, because the potential of the hydrogen half-cell is zero. Of course, we can do the same thing for every other half-cell, but we need not do so, since the hydrogen/hydrogen ion half-cell is hard to work with because it involves a gas. We can get an unknown half-cell potential from its cell potential with any half-cell of known potential. For example, once we get the copper half-cell potential, we can use it to calculate the unknown zinc halfcell potential from the Daniell cell potential. A collection of half-cell potentials, all written as reductions, is presented as Table 17.2. 

Test method for porosity in gold platings on metal substrates by gas exposures Test method for half-cell potentials of uncoated reinforcing steel in concrete Method for detection of copper corrosion from petroleum products by the copper strip tarnish test... 

Suppose the question is whether silver will be oxidized if it is immersed in copper sulfate. The half-cell potential for Ag-Ag+ is —0.80 volt and that for Cu-Cu+2 is —0.34 volt. The first value, —0.80 volt, is more negative than the second, —0.34 volt. The difference, then, is still negative —0.80 — (—0.34) = —0.46 volt. The negative answer shows that Ag-Ag+ has less tendency to lose electrons than does Cu-Cu+2. The reaction will not tend to proceed spontaneously. Silver will not be oxidized to an appreciable extent in copper sulfate. 

The three values of ° are easily calculated from half-cell potentials. Then, we can predict with confidence that reaction (65) will not occur to an appreciable extent if solid copper is immersed in dilute acid. The negative value of ° (—0.34 volt) indicates that equilibrium in (65) strongly favors the reactants, not the products. 

Then add the copper half-cell reduction to the zinc half-cell oxidation and add the half-cell potentials ... 

Worked Example 3.17. Consider the simple cell SCE 11 Cujj j Cu(s> . From a measurement of the emf and a knowledge of sce. die electrode potential, cu2+,Cu> was determined to be 0.300 V. We will consider three possible situations, and calculate the activity of the copper ion in each case (i) Ej = 0, (ii) Ej = 30 mV, with the junction adding to the electrode potential for the copper half cell, and (iii) Ej = 30 mV, with the junction subtracting from the electrode potential for the copper half cell. As usual, we will assume a value of a(Cu) = 1 throughout and E 2+ = 0.340 V. 

Since this potential is positive, tthe reaction will proceed to the right electrons will be withdrawn from the copper electrode and flow through the external circuit into the silver electrode. Note carefully that in combining these half-cell potentials, we did not multiply E° the for the Cu2+/Cu couple by two. The reason for this will be explained later. 

Given the half-cell potentials in Table 17.2, calculate the cell potential of (a) the Daniell cell, (b) the copper/silver cell. 

In a diagram of a cell a single vertical line conventionally represents a phase boundary at which a potential difference is taken into account. A double vertical line represents a liquid junction at which the potential difference is ignored or is considered to be eliminated by an appropriate salt bridge. For example, a cell consisting of zinc and copper half-cells can be expressed by... 

The more positive electrochemical values favor the formation of reduced species. For example, the Nemst equation predicts that thorium with a half cell potential of E Th = -1.9 V, would have to be at a concentration of over 10 M before it would plate out at the half-cell potential of copper, E° Cu = 0.34 V. Even when a reverse pulse plating process is employed to create level, and polycrystalline, copper, the Nemst equation predicts thorium would still have to be at a concentration of over 10 M before it would plate out at -0.34 V. One would expect to obtain extreme copper purity from contaminants such as Th when electroplating at the voltages required for copper plating. 

We have also pursued electrochemically back-plating of the copper sample to reduce the copper ion concentration and leave in solution impurities such as thorium and uranium, which should not plate out at the half-cell potential of copper. Theoretically, the amount of sample that can be processed in this manner is not limited. All materials including any non-sample electrodes must not add contamination and must be of extreme purity. Also, the amount of copper remaining in solution must be back-plated to <10 (xg/ml, and if a sulfate system is used, which is useful in support of further developing the predictive rejection rate information, then the sulfate ion should be <10 mmol as well. This approach hinges on the rejection rate remaining sufficiently high as to not introduce an undue amount of error. We have measured rejection rates as low as 10 but even at 10 this would only represent a 1% error in the assay result. 

Figure 21-6 describes this potential calculation. The graph shows the zinc half-cell with the lower reduction potential (the oxidation half-reaction) and the copper half-cell with the higher reduction potential (the reduction halfreaction). You can see that the space between the two (Fceii) difference between the potentials of the individual half-cells. The Example Problem that follows gives a step-by-step description of calculating cell potentials. 

If the zinc and copper half-cells are combined, the copper half-cell will be the cathode because it has the more positive half-cell reduction potential. The galvanic cell voltage under standard-state conditions (Fig. 17.4) will be... 

The SHE functions as the anode in this cell, and Cu ions oxidize H2 to H+ ions. The standard electrode potential of the copper half-cell is 0.337 volt as a cathode in the Cu-SHE cell. 

As we have seen, different concentrations of ions in a half-cell result in different half-cell potentials. We can use this idea to construct a concentration cell, in which both half-cells are composed of the same species, but in different ion concentrations. Suppose we set up such a cell using the Cu +/Cu half-cell that we introduced in Section 21-9. We put copper electrodes into two aqueous solutions, one that is 0.10 M CUSO4 and another that is 1.00 M CUSO4. To complete the cell construction, we connect the two electrodes with a wire and join the two solutions with a salt bridge as usual (Figure 21-14). Now the relevant reduction half-reaction in either half-cell is... 

If copper is placed in the solution of pH = 8, then calculation of the half-cell potential of the copper will depend on the copper-ion concentration of the solution. Generally, this will be small and unknown. For purposes of estimation, it may be assumed that aCu2+ = 10-4 = mCu2+, and that if corrosion occurs, the cathodic reaction is the release of hydrogen gas at one atmosphere pressure (solution is deaerated). The reaction under consideration is then ... 

The metals must have sufficiently positive half-cell potentials that corrosive reactions that would change the potential to a corrosion potential, Ecorr, do not occur. This requirement, with few exceptions, restricts the metal component of the half cell to silver, mercury, and copper. For these metals, appearing in Table 6.1, corrosion due to hydrogen evolution will not occur since the metal half-cell potentials are above potentials for hydrogen evolution. Also, the kinetics of the reduction of any dissolved oxygen are sufficiently slow that the potential is shifted negligibly from that of the metal half cell. 

Calculate the standard potential of the cell produced when the copper(ll)/copper half-cell (e° = 0.34 V) is combined with the iron(III)/iron(ll) half-cell (e° = 0.77 V). 

Determining Ehaif-ceii The Standard Hydrogen Electrode What portion of ceii for the zinc-copper reaction is contributed by the anode half-cell (oxidation of Zn) and what portion by the cathode half-cell (reduction of Cu ) That is, how can we know half-cell potentials if we can only measure the potential of the complete cell Half-cell potentials, such as Ezine and °opper. are not absolute quantities, but rather are values relative to that of a standard. This standard reference halfcell has its standard electrode potential defined as zero (E fereiice — 0.00 V). The standard reference half-cell is a standard hydrogen electrode, which consists of a specially prepared platinum electrode immersed in a 1 M aqueous solution of a strong acid, H (fl ) [or H30 (a )], through which H2 gas at 1 atm is bubbled. Thus, the reference half-reaction is... 

The graph in Figure 20.7 shows how the zinc half-cell with the lower reduction potential and the copper half-cell with the higher reduction potential are related. 

For example, if a hydrogen half cell is connected to a copper half-cell, (i.e. a copper rod in a solution of copper ions) and the potential of the cell measured, it is found to be 4-0.34 volts, and assigned as the electrode potential of the copper half cell, using the following convention. If the metal electrode loses electrons more readily than the hydrogen electrode, the metal eventually acquires a negative potential due to the accumulation of electrons on the metal. If the metal electrode does not form ions as readily as the hy-... 

Zinc metal is oxidized to zinc ions, and copper ions are reduced to copper metal. The copper cathode becomes depleted of electrons because these are taken up by the copper ions in solution. At the same time the zinc anode has an excess of electrons because the neutral zinc atoms are becoming ionic and liberating electrons in the process. The excess electrons from the anode flow to the cathode. The flow of electrons is the source of external current the buildup of electrons at the anode and the depletion at the cathode constitute a potential difference that persists until the reaction ceases. The reaction comes to an end when either all the copper ions are exhausted from the system or all the zinc metal is dissolved, or an equilibrium situation is reached when both half-cell potentials are equal. If the process is used as a battery, such as a flashlight battery, the battery becomes dead when the reaction ceases. 

We can always change the terminal of the voltmeter to which we connect an electrode, so for the time being let s simply adjust the measurement to show a positive voltage every time. If we apply this technique to the example of copper and silver, what do we observe The cell Cu(s) Cu (l M) Ag (l M) Ag(s) has a potential of 0.462 V. If we take the same copper half-cell and connect it instead to a half-cell that reduces iron(III) to iron(II), we find that the measured cell potential is 0.434 V. If this iron electrode is connected to the silver one, the resulting cell potential is 0.028 V. These numbers are clearly related 0.462 = 0.434 + 0.028 (Figure 13.8). This observation is important for two reasons. First, it shows a behavior of cell potentials that is akin to that of a state function. Second, it suggests that if we choose a specific standard electrode to which we compare all other electrodes, we can devise a practical system for determining cell potential. 

Let s calculate the potential of the cell in Figure 12.2. By convention 5, both half-cells are written and calculated as reductions, no matter what is happening in the real cell. Thus, for the copper half-cell,... 

The cell emf is the sum of the potentials for the reduction and oxidation halfreactions. For the cell we have been describing, the emf is the sum of (he reduction potential (electrode potential) for the copper half-cell and the oxidation potential (negative of the electrode potential) for the zinc half-cell. 

A voltaic cell is made from a Co (aq)/Co(s) halfcell and a Cu (aq)/Cu(s) half-cell. The cell potential was +0.62 V, with the copper half-cell positive. Calculate the standard electrode potential of the cobalt half-cell. 

---