The standard electrode potentials of half-cells can be arranged in numerical order. This list is known as the electrochemical series. By organizing half-cells in this way, we can easily compare their relative abilities to gain or lose electrons. This allows us to predict the direction of redox reactions and calculate the potential difference of combined electrochemical cells.
A list of standard electrode potentials arranged in numerical order from the most negative (most reducing) to the most positive (most oxidising).
🔑 Key Principle
The half-cell with the more negative standard electrode potential has a stronger tendency to lose electrons, meaning it undergoes oxidation. The half-cell with the more positive standard electrode potential has a stronger tendency to gain electrons, meaning it undergoes reduction.
Arrangement of the Series
By convention, standard electrode potentials are written as reductions, with the most negative potentials placed at the top of the table. A representation of the electrochemical series is shown below:
A substance that donates electrons to another species, reducing them while itself being oxidised. Strong reducing agents are found on the right-hand side of half-equations with very negative potentials.
A substance that accepts electrons from another species, oxidising them while itself being reduced. Strong oxidising agents are found on the left-hand side of half-equations with very positive potentials.
Calculating Standard Cell Potentials (\( E^\theta_{\text{cell}} \))
When two half-cells are combined to form a full electrochemical cell, a potential difference is generated. We calculate the standard electromotive force (EMF) or cell potential (\( E^\theta_{\text{cell}} \)) using the standard potentials of the reduction and oxidation half-cells:
\[ E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} \]
Alternatively, since reduction always occurs at the positive electrode (cathode) and oxidation at the negative electrode (anode):
\[ E^\theta_{\text{cell}} = E^\theta_{\text{positive}} - E^\theta_{\text{negative}} \]
Because the potential of a cell is a physical measurement of energy per charge, you must never multiply the standard electrode potential values by stoichiometric coefficients, even if the overall balanced cell equation requires balancing the electrons!
Predicting the Feasibility of Redox Reactions
We can use standard electrode potentials to determine whether a proposed redox reaction is thermodynamically feasible. For a reaction to proceed spontaneously under standard conditions:
- The overall cell potential \( E^\theta_{\text{cell}} \) must be greater than zero (\( E^\theta_{\text{cell}} > 0 \)).
- The species undergoing reduction must have the more positive standard electrode potential.
- The species undergoing oxidation must have the more negative standard electrode potential.
Standard cell potential predictions assume standard conditions of 1.00 mol dm^-3 concentrations, 100 kPa pressure, and 298 K temperature. If the actual concentrations of reactants are changed, the position of equilibrium in the half-cells shifts. This changes the actual electrode potentials, which can make a reaction feasible even if its standard cell potential is negative, or vice versa. Always check if the question mentions non-standard conditions!
Thermodynamic vs. Kinetic Feasibility
Just because standard electrode potentials predict that a reaction is feasible, it does not mean the reaction will occur rapidly, or even visibly, in practice. There are two main reasons for this:
- High Activation Energy: The reaction may have a very high activation energy barrier. Even though the reaction is thermodynamically spontaneous, its rate is so slow that it is kinetically stable.
- Non-Standard Conditions: As noted above, if the concentrations, temperature, or pressure are different from standard values, the actual cell potential will differ from the calculated standard value.
1. \( \text{Fe}^{3+}\text{(aq)} + \text{e}^- \rightleftharpoons \text{Fe}^{2+}\text{(aq)} \quad E^\theta = +0.77\text{ V} \)
2. \( \text{I}_2\text{(s)} + 2\text{e}^- \rightleftharpoons 2\text{I}^-\text{(aq)} \quad E^\theta = +0.54\text{ V} \)
Calculate the standard cell potential for a cell combining these half-cells, write the overall balanced equation, and state whether the reaction between iron(III) ions and iodide ions is feasible.
Step 1: Identify which half-cell undergoes reduction and which undergoes oxidation:
The iron(III)/iron(II) half-cell has the more positive potential (+0.77 V compared to +0.54 V). Therefore, the iron(III) half-cell will undergo reduction (gaining electrons):
\( \text{Fe}^{3+}\text{(aq)} + \text{e}^- \rightarrow \text{Fe}^{2+}\text{(aq)} \)
The iodine/iodide half-cell has the more negative potential. Therefore, it will undergo oxidation (losing electrons):
\( 2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)} + 2\text{e}^- \)
Step 2: Combine the half-equations into a balanced equation:
Multiply the reduction half-equation by 2 to balance the electrons:
\( 2\text{Fe}^{3+}\text{(aq)} + 2\text{e}^- \rightarrow 2\text{Fe}^{2+}\text{(aq)} \)
Add the equations together to get the overall equation:
\( 2\text{Fe}^{3+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + \text{I}_2\text{(s)} \)
Step 3: Calculate the standard cell potential:
\[ E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} \]
\[ E^\theta_{\text{cell}} = (+0.77\text{ V}) - (+0.54\text{ V}) = +0.23\text{ V} \]
Since \( E^\theta_{\text{cell}} > 0 \), the forward reaction between iron(III) and iodide ions is feasible under standard conditions.
Standard potentials:
\( \text{Cl}_2\text{(g)} + 2\text{e}^- \rightleftharpoons 2\text{Cl}^-\text{(aq)} \quad E^\theta = +1.36\text{ V} \)
\( \text{Br}_2\text{(l)} + 2\text{e}^- \rightleftharpoons 2\text{Br}^-\text{(aq)} \quad E^\theta = +1.09\text{ V} \)
Step 1: Write down the proposed reduction and oxidation reactions:
The question asks if chlorine will oxidise bromide ions. This means chlorine gas (\( \text{Cl}_2 \)) would undergo reduction:
\( \text{Cl}_2\text{(g)} + 2\text{e}^- \rightarrow 2\text{Cl}^-\text{(aq)} \)
And bromide ions (\( \text{Br}^- \)) would undergo oxidation:
\( 2\text{Br}^-\text{(aq)} \rightarrow \text{Br}_2\text{(l)} + 2\text{e}^- \)
Step 2: Calculate the cell potential for this combination:
Reduction is chlorine (+1.36 V) and oxidation is bromide (+1.09 V):
\[ E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} \]
\[ E^\theta_{\text{cell}} = (+1.36\text{ V}) - (+1.09\text{ V}) = +0.27\text{ V} \]
Step 3: State the conclusion:
Since the cell potential is positive (+0.27 V), the reaction is feasible. Chlorine gas will spontaneously oxidise bromide ions under standard conditions.
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