Question

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s) \\ \text { (b) } \mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q) \\ \text { (c) } \mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q) \\ \text { (d) } 2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow \\ 2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{Cu}^{2+}(a q) \end{array} $$

Short answer

The standard emfs for the given reactions are: (a) \(+0.82 \,V\) (b) \(-1.37 \,V\) (c) \(+1.21 \,V\) (d) \(+0.80 \,V\)

Step-by-step solution

Identify the half-reactions for each reaction

For each given reaction, we need to identify the reduction and oxidation half-reactions. (a) Cl2(g) + 2I^-(aq) -> 2Cl^-(aq) + I2(s) Reduction half-reaction: Cl2(g) + 2e^- -> 2Cl^-(aq) Oxidation half-reaction: 2I^-(aq) -> I2(s) + 2e^- (b) Ni(s)+2Ce^4+(aq) ->Ni^2+(aq)+2Ce^3+(aq) Reduction half-reaction: 2Ce^4+(aq) + 2e^- -> 2Ce^3+(aq) Oxidation half-reaction: Ni(s) -> Ni^2+(aq) + 2e^- (c) Fe(s)+2Fe^3+(aq) -> 3Fe^2+(aq) Reduction half-reaction: 2Fe^3+(aq) + 6e^- -> 2Fe^2+(aq) Oxidation half-reaction: 4Fe(s) -> 4Fe^2+(aq) + 8e^- (d) 2NO3^-(aq) + 8H^+(aq) + 3Cu(s) -> 2NO(g)+4H2O(l)+3Cu^2+(aq) Reduction half-reaction: 2NO3^-(aq) + 12H^+(aq) + 10e^- -> 2NO(g) + 6H2O(l) Oxidation half-reaction: 3Cu(s) -> 3Cu^2+(aq) + 6e^-

Look up standard reduction potentials

Using Appendix E, we look up the standard reduction potentials (E°) for the half-reactions. (a) E°(reduction) = +1.36 V (for Cl2 + 2e^- -> 2Cl^-) E°(oxidation) = +0.54 V (for I2 + 2e^- -> 2I^-) (b) E°(reduction) = -1.61 V (for Ce^4+ + e^- -> Ce^3+) E°(oxidation) = -0.24 V (for Ni^2+ + 2e^- -> Ni) (c) E°(reduction) = +0.77 V (for Fe^3+ + 3e^- -> Fe) E°(oxidation) = -0.44 V (for Fe^2+ + 2e^- -> Fe) (d) E°(reduction) = +0.96 V (for NO3^- + 6H^+ + 3e^- -> NO + 3H2O) E°(oxidation) = +0.16 V (for Cu^2+ + 2e^- -> Cu)

Calculate the standard emf for each reaction

To calculate the standard emf (E°) for each reaction, subtract the standard reduction potential of the oxidation half-reaction from the reduction half-reaction. (a) E°(reaction) = E°(reduction) - E°(oxidation) = 1.36 V - 0.54 V = +0.82 V (b) E°(reaction) = E°(reduction) - E°(oxidation) = -1.61 V - (-0.24 V) = -1.37 V (c) E°(reaction) = E°(reduction) - E°(oxidation) = 0.77 V - (-0.44 V) = +1.21 V (d) E°(reaction) = E°(reduction) - E°(oxidation) = 0.96 V - 0.16 V = +0.80 V So, the standard emfs for the given reactions are: (a) +0.82 V (b) -1.37 V (c) +1.21 V (d) +0.80 V

Key concepts

Standard Reduction Potentials

Standard reduction potentials are crucial in electrochemistry. They help in predicting the direction of redox reactions and calculating the voltage of electrochemical cells.
These potentials are measured in volts, often found in tables like Appendix E. The values indicate how easily a substance is reduced.
A more positive value means a greater tendency to gain electrons (get reduced).

• To use these potentials, identify the reduction and oxidation half-reactions in your full redox equation.
• Each half-reaction has a standard reduction potential.
• By convention, reduction potentials are always given as they involve the gain of electrons.

When calculating the emf (electromotive force) of a cell, standard reduction potentials for oxidation reactions must be reversed and their signs changed.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are at the heart of electrochemistry. These reactions involve the transfer of electrons between substances, changing their oxidation states. Understanding them is key to mastering many chemical processes. An oxidation reaction loses electrons, and a reduction reaction gains them:

• **Oxidation**: Loss of electrons, increase in oxidation state.
• **Reduction**: Gain of electrons, decrease in oxidation state.

In any redox reaction, there will always be a substance that gets oxidized and one that gets reduced. They're connected by the flow of electrons between them. Identifying the half-reactions helps in using standard reduction potentials effectively. Pay attention to balancing these reactions, as the electrons lost and gained must be equal.

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy through redox reactions. These cells are foundational in batteries and various chemical sensors. There are two main types: galvanic (or voltaic) and electrolytic cells.
Galvanic cells are spontaneous and generate electricity. Electrolytic cells require external power to drive non-spontaneous reactions. Key parts of an electrochemical cell include:

• **Anode**: Site of oxidation, releases electrons.
• **Cathode**: Site of reduction, gains electrons.
• **Salt Bridge** or **Porous Disk**: Allows ions to flow while keeping the solutions in the two half-cells neutral overall.

The cell potential ( E° ) is calculated from standard reduction potentials. It determines a cell's ability to produce an electric current. Calculating the potential involves the difference between the cathode and anode potentials. A positive cell potential suggests a spontaneous redox reaction.