Chapter 9: Electrochemistry (C8906973)

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Electrochemistry is the study of electricity in the context of chemical reactions, specifically, the transfer of electrons. This makes sense because electricity is the flow of charge (from ), and electrons are charged particles. Chemical reactions involving transfer of electrons are known as oxidation/reduction, or redox reactions. The electron acceptor is itself reduced, and the electron donor is itself oxidized. This can be memorized with the mnemonic “OIL RIG”, which states that “oxidation is loss of electrons, and reduction is gain of electrons”.

To study the transfer of electrons, it is necessary to know where the electrons are, which are indicated by the oxidation state of specific atoms in a chemical compound. Oxidation state defines the hypothetical charge an atom would have if all bonds in the compound were 100% ionic. Note that this charge is hypothetical, and the atom may or may not have this charge. The rules for assigning oxidation state are ordered by importance of precedence, which from the greatest to the least veracity, including:

  1. Free elements (which are not bound to elements apart from itself, and therefore has no charge) have an oxidation state of 0. For example, the hydrogens in [latex]\ce{H2}[/latex], or oxygens in [latex]\ce{O2}[/latex] [both useful in outlining the importance of precedence of rules, when for example, comparing against the next rule, and the one after, respectively], have an oxidation state of 0
  2. The summation of the oxidation states of a compound or ion, must equal to the net charge on that compound or ion. For example, in [latex]\ce{SO4^{2-}}[/latex], the oxygens have an oxidation state of -2 (later rule), and so [latex]S+4(-2)=-2[/latex], or solving, [latex]S=+6[/latex], meaning the sulfur atom has an oxidation state of [latex]+6[/latex]
  3. Hydrogen has an oxidation state of [latex]+1[/latex], except when bonded with a metal, when it has an oxidation state of [latex]-1[/latex]. For example, the hydrogens in [latex]\ce{H2O}[/latex], or even [latex]\ce{H2O2}[/latex] [again, useful in outlining the importance of precedence of rules, when for example, comparing against the next rule], have an oxidation state of +1. In contrast, the hydrogens in NaH has an oxidation state of -1
  4. Oxygen has an oxidation state of -2. For example, the oxygen in [latex]\ce{H2O}[/latex] has an oxidation state of -2

Other rules which you may come across include:

  • Group Ia elements have an oxidation state of +1, when in compound
  • Group IIa elements have an oxidation state of +2, when in compound
  • Aluminum has an oxidation state of +3, when in compound
  • Fluorine has an oxidation of -1, when in compound

An example of a redox reaction is [latex]\ce{2H2+O2 -> 2H2O}[/latex], which is the combustion of hydrogen fuel, used in rockets. By the rules, note that the hydrogens in [latex]\ce{H2}[/latex] and oxygens in [latex]\ce{O2}[/latex] both have an oxidation state of 0. In the compound [latex]\ce{H2O}[/latex], the hydrogen has an oxidation state of +1, and the oxygens have an oxidation state of -2. Therefore, hydrogen is oxidized, and oxygen is reduced. By the law of conservation of charge, oxidation and reduction occurs together. Because hydrogen is oxidized, the other substance, oxygen, is the oxidizing agent (aka oxidant). Analogously, because oxygen is reduced, the other substance, hydrogen, is the reducing agent (aka reductant). It is incorrect to say that what is oxidized and reduced, are the reducing agent and oxidizing agent respectively, because whereas atoms [in the compound] are oxidized and reduced, the compound itself is the reducing and oxidizing agents.

Like acid-base titration, there is redox titration, which is the use of redox reactions to find unknown concentrations. Salt bridges can be used to connect two oxidation and reduction half-cells [of a Galvanic cell, discussed ], to permit the flow of electron from one half-cell to another, which is possible because there is a potential difference between the half-cells. This potential difference can be measured with a voltmeter. By adding a reductant or oxidant, the potential difference can thus be altered. For example, if there is a solution with an unknown concentration of [latex]\ce{Sn2+}[/latex] ions, given the equation [latex]\ce{Sn2+ + I3^- -> Sn^{4+} + 3I^{-} }[/latex], it is titrated against a known concentration of [latex]I3^-[/latex]. Since every [latex]I3^-[/latex] molecule oxidizes one [latex]Sn^{2+}[/latex] ion into [latex]Sn^{4+}[/latex]. As the [latex]Sn^{2+}[/latex], the potential difference [on the y-axis], plotted against the volume of the titrant added [on the x-axis], creates the familiar sigmoidal titration curve. At the equivalence point, [latex]Sn^{2+}[/latex] is completely oxidized into [latex]Sn^{4+}[/latex].

Redox reactions can be separated into two half reactions, specifically, the (1) reduction and (2) oxidation reactions. Each of these half reactions, have an associated electrochemical potential, as electrons are being transferred. Note that the half reactions don’t occur in isolation however, since both halves are required for a redox reaction to occur. Note also, that the potential required to oxidize [latex]X -> X^+[/latex], is the exact opposite [so same magnitude, but opposite sign] to the potential required to reduce [latex]X^+ -> X[/latex], so only one listing is required, usually, the reduction half-reaction potential. Since there is no absolute zero potential, the reduction of [latex]\ce{H+}[/latex] has been arbitrarily assigned a standard electrode potential ([latex]E^{\circ}[/latex]) of 0V, namely, [latex]\ce{2H+  + 2e- \leftrightharpoons H_{2(g)} }[/latex], and the others measured by this standard. The words “standard” precede the name, because the potential varies under different conditions, it is given in standard state (discussed ), namely, at [latex]25^{\circ}[/latex], 1M concentration of reactants and products in solution, and at 1 atm for gases. Electrochemical potential is an intensive property, meaning it is a property that doesn’t depend on quantity (discussed ), for example, [latex]\ce{Sn^{4+} + 2e- \leftrightharpoons Sn^{2+} }[/latex] occurs at a reduction potential of +0.15V. Even if the entire equation were doubled, this would not alter the reduction potential of +0.15V. It is notable that the money metals have positive reduction potential [and therefore negative oxidation potential], including platinum, gold, silver, copper and mercury. Nickel is the exception, with a slightly negative reduction potential [and therefore positive oxidation potential]. A negative oxidation potential indicates that the oxidation is non-spontaneous, which makes sense that the money metals don’t oxidize easily.

1 Electrolytic cell

Electrolytic cell is where a power source is applied across a Galvanic cell (discussed ) to force current in one direction. Unlike the Galvanic cell, electrolytic cells may not have a positive cell potential, and often doesn’t. Additionally, electrolytic cells do not require a salt bridge, instead, often, both electrodes sharing the same electrolyte. Like Galvanic cells, oxidation still takes place at the anode, and reduction at the cathode. The difference however, is that the anode in the electrolytic cell is positive, and the cathode is negative [i.e. flipped around the other way from as it was in the Galvanic cell]. Unlike Galvanic cells, electrolytic cells do not generate a current [as current is being applied]. These two facts can be a mnemonic for the signs of the electrodes, because the Galvanic cell focuses on electricity production [and thus the electrical circuit], and electricity is the flow of electrons, the electrons move from the negative anode toward the positive cathode. In contrast, in electrolytic cells, the focus is on the solution, and cations move toward the cathode. Since cations are positive, it moves towards a negative cathode. Rather, it is used in electrophoresis (discussed ), metal plating, and purification of metals. Metal cations move towards the cathode, and become solid metal, either plating the cathode, or if made of the same metal, grows into a larger piece of pure metal.

2 Galvanic (voltaic) cell

As redox reaction has potentials associated with it, there is potential for the reaction to do useful work. This can occur in a galvanic (aka voltaic) cell, which is the concept behind batteries. Galvanic cell separates the half reactions in separate solutions, known as half cells. The two solutions are connected by a salt bridge, which permits ions to transfer between the solutions. The half reactions take place at the electrodes in the half cells, the electrode where reduction occurs known as the cathode, which is positive; and the electrode where oxidation occurs known as the anode, which is negative. The mnemonic to remember this is “REDCAT” and Rico [from Hannah Montana]’s famous “Hey-ohhhh”, which sounds like “AO”, for “reduction at the cathode, and anode is oxidation”. The [voltaic] cell can then generate a current, as electrons transfer from the solution in one half cell to the other. The sum of the two half potentials, is known as the electromotive force (emf, aka cell potential [latex]E[/latex]), which is not a force, but a voltage. In a Galvanic cell, the cell potential must be positive.

An example of a Galvanic cell is a zinc anode, and a copper cathode. Remember that copper is a money metal, so it is less likely to oxidize. The zinc anode is oxidized [to lose 2 electrons] to form zinc ions, and the copper cathode is reduced [to gain 2 electrons] to form copper metal. The standard electrode potential of [latex]Zn -> Zn^{2+}[/latex], is the opposite of [latex]Zn^{2+} + 2e^- -> Zn_{s}[/latex], and is therefore +0.76V (rather than the usual minus for the reduction equivalent). The standard electrode potential of [latex]Cu^{2+} + 2e^- \rightleftharpoons Cu_{(s)}[/latex], is +0.34V. The cell potential is therefore [latex]0.76+0.34=1.1V[/latex]. The positive emf means it is voltaic. Despite the overall equation [latex]\ce{Zn+CuSO4 -> ZnSO4 + Cu}[/latex] is already balanced, even if it were not, because emf is an intensive property (explained ), this is unnecessary.

In contrast, Gibbs free energy [of a cell] will require balancing the equation. It can be found with [latex]\Delta G=-nFE[/latex], where [latex]\Delta G[/latex] is the change in Gibbs free energy, [latex]n[/latex] is the number of electrons transferred [when the equation is balanced], [latex]F[/latex] is Faraday’s constant (which is the magnitude of electric charge on 1 mol of electrons, namely, [latex]9.65\times 10^4 C/mol[/latex], and [latex]E[/latex] is the cell’s EMF. Where the cell is at standard state conditions, the equation can be rewritten as [latex]\Delta G^{\circ}=-nF.E^{\circ}[/latex], where [latex]\Delta G^{\circ}[/latex] is the free energy change starting from standard state conditions. This is advantageous because [latex]\Delta G^{\circ}[/latex] is indexed. The relationship between [latex]\Delta G[/latex] and [latex]\Delta G^{\circ}[/latex] is [latex]\Delta G=\Delta G^{\circ} + RT.ln{Q}[/latex], where [latex]R[/latex] is the universal gas constant, [latex]T[/latex] is the [absolute] temperature, and [latex]Q[/latex] is the reaction quotient (used to predict the direction a chemical reaction will move to reach equilibrium, discussed ). Remember (from ) that Gibbs free energy is the energy available to do non-PV work. When the chemical reactions taking place inside a battery have reached equilibrium, they can do no work, as the battery is “dead”, meaning [latex]\Delta G=0[/latex]. When the reaction is at equilibrium, [latex]Q=K[/latex], as stated , where [latex]K[/latex] is the equilibrium constant. In-substituting, and reshuffling, [latex]\Delta G^{\circ}=-RT.ln{K}[/latex], meaning that if [latex]\Delta G^{\circ}[/latex] is positive (i.e. non-spontaneous reaction), [latex]K<1[/latex]; if [latex]\Delta G^{\circ}[/latex] is negative (i.e. spontaneous reaction), [latex]K>1[/latex]; and if [latex]\Delta G^{\circ}=0[/latex], [latex]K=1[/latex].

Since it is known that [latex]\Delta G=\Delta G^{\circ} + RT.ln{Q}[/latex], [latex]\Delta G=-nFE[/latex], and [latex]\Delta G^{\circ}=-nF.E^{\circ}[/latex] (from ), in-substituting, [latex]-nFE=-nF.E^{\circ}+RT.ln{Q}[/latex], or reshuffling, [latex]E=E^{\circ}-\dfrac{RT}{nF}.ln{Q}[/latex]. Given the standard conditions of [latex]T=25^{\circ}C=298.15K[/latex], [latex]R=8.3J/K/mol[/latex], and inserting Faraday’s constant, the equation becomes [latex]E=E^{\circ}-\dfrac{(8.3)(298.15)}{n\times(9.65\times 10^4)}.ln{Q}=E^{\circ}-\dfrac{0.026}{n}.ln{Q}[/latex]. Since the change of base formula is [latex]log_{b}{x}=\dfrac{log_{k}{x} }{log_{k}{b} } [/latex], in-substituting to convert [latex]ln{Q}[/latex], [latex]log_{e}{x}=\dfrac{log_{10}{x} }{log_{10}{e} }=2.3\times log{x}[/latex]. Therefore, back-substituting, [latex]E=E^{\circ}-\dfrac{0.06}{n}.log{Q}[/latex], known as the Nernst equation. The Nernst equation relates chemical concentration, with electrical potential, and is used to find cell potential of cells not at standard state condition.


Line notation for Galvanic cells include [going from left to right] the anode substance written first, ions formed by that substance, then ions formed by reactant at cathode, and then the cathode substance. Effectively, the electrodes are outside, and the ions produced are on the inside; and the anode is found on the left. So this may look like [latex]\ce{Mg | Mg2+ || Al3+ | Al}[/latex].The electrodes and ions formed are separated by a single line, and the half cells are separated by a double line.

Concentration cell is a type of galvanic cell, which has two half-cells running the same reaction, but in opposite directions, due to differences in concentrations of the two half-cells. For example, the anode may be [latex]\ce{Cu_{(s)} -> Cu^{2+} + 2e^- }[/latex], and the cathode is thus [latex]\ce{Cu^{2+} + 2e^- -> Cu_{(s)} }[/latex], giving a full equation of [latex]\ce{Cu^{2+}_{(cathode)} + Cu_{(s)} + 2e^- -> Cu^{2+}_{(anode)} + Cu_{(s)} + 2e^-}[/latex]. The reaction runs because of the desire to converge the concentrations in the two cells; the desire to increase entropy. Because anode is associated with oxidation [which is loss of electron, aka gain of proton], it will be the side with fewer [latex]Cu^{2+}[/latex] ions [so it can gain more protons]. Evidently, because the concentrations of the two half-cells are distinct, they cannot be both at the standard state of 1M concentration. As such, the Nernst equation is required, [latex]E=E^{\circ}-\dfrac{0.06}{n}.log{Q}[/latex]. Remember that at standard state, because the concentrations are the same, there is no electrical potential, meaning [latex]E^{\circ}=0[/latex]. There are two electrons in the equation, meaning [latex]n=2[/latex], and the reaction quotient [latex]Q=\dfrac{Cu^{2+}_{(anode)} }{Cu^{2+}_{(cathode)} }[/latex]. Since [latex]log(\dfrac{a}{b})=log{a}-log{b}=-(log{b}+log{a})=-log(\dfrac{b}{a})[/latex], note that even if the reaction quotient has been flipped around, the result will still be the same, except with a negative cell EMF, which is not possible for a Galvanic cell.

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General chemistry - Pre-med science - MR. SHUM'S CLASSROOM