A brief note on Redox and Electrode Potentials . Read it before a test. Try each question at the end before you open its answer.
Half-cells and the Daniell cell A half-cell is a metal in a solution of its ions. A couple is the oxidised and reduced forms together, such as Zn 2 + / Zn \text{Zn}^{2+}/\text{Zn} Zn 2 + / Zn . Anode: oxidation, negative. Cathode: reduction, positive. Electrons flow from anode to cathode in the wire. Ions carry the charge through the salt bridge. Cell diagram: Zn ( s ) ∣ Zn 2 + ( a q ) ∥ Cu 2 + ( a q ) ∣ Cu ( s ) \text{Zn}(s) \mid \text{Zn}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu}(s) Zn ( s ) ∣ Zn 2 + ( a q ) ∥ Cu 2 + ( a q ) ∣ Cu ( s ) , with the anode on the left. Faraday: Q = I t Q = It Q = I t and one mole of electrons carries 96500 C 96500\ \text{C} 96500 C . Electrode potentials The standard hydrogen electrode is 0.00 V 0.00\ \text{V} 0.00 V : platinum, H + \text{H}^+ H + at 1 mol/L 1\ \text{mol/L} 1 mol/L , H 2 \text{H}_2 H 2 at 1 bar 1\ \text{bar} 1 bar . E ∘ E^\circ E ∘ is a reduction potential at 1 mol/L 1\ \text{mol/L} 1 mol/L , 1 bar 1\ \text{bar} 1 bar and 298 K 298\ \text{K} 298 K . E c e l l ∘ = E c a t h o d e ∘ − E a n o d e ∘ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} E ce l l ∘ = E c a t h o d e ∘ − E an o d e ∘ . Reverse a half-reaction: change the sign. Multiply it: no change. Predicting reactions Lower E ∘ E^\circ E ∘ : stronger reducing agent. Higher E ∘ E^\circ E ∘ : stronger oxidising agent. A reaction can happen if E ∘ ( reduced ) − E ∘ ( oxidised ) E^\circ(\text{reduced}) - E^\circ(\text{oxidised}) E ∘ ( reduced ) − E ∘ ( oxidised ) is positive. Metals with E ∘ < 0 E^\circ < 0 E ∘ < 0 release hydrogen from dilute acid. Δ G ∘ = − n F E ∘ \Delta G^\circ = -nFE^\circ Δ G ∘ = − n F E ∘ . E ∘ E^\circ E ∘ says whether a reaction can happen, not how fast. A worked example A cell has a zinc anode and a silver cathode. Find its potential and Δ G ∘ \Delta G^\circ Δ G ∘ , for the reaction Zn + 2 Ag + → Zn 2 + + 2 Ag \text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag} Zn + 2 Ag + → Zn 2 + + 2 Ag .
Potential. E c e l l ∘ = 0.80 − ( − 0.76 ) = 1.56 V E^\circ_{cell} = 0.80 - (-0.76) = 1.56\ \text{V} E ce l l ∘ = 0.80 − ( − 0.76 ) = 1.56 V . Electrons. The equation moves n = 2 n = 2 n = 2 . Energy. Δ G ∘ = − 2 × 96500 × 1.56 = − 301 080 J/mol \Delta G^\circ = -2 \times 96500 \times 1.56 = -301\,080\ \text{J/mol} Δ G ∘ = − 2 × 96500 × 1.56 = − 301 080 J/mol , or − 301.1 kJ/mol -301.1\ \text{kJ/mol} − 301.1 kJ/mol . Practice with new numbers
A program to try The program finds the cell potential and the free energy change of three cells.
Run it. Which cell gives out the most energy?
1. Add a cell with cathode + 0.34 V +0.34\ \text{V} + 0.34 V and anode − 0.13 V -0.13\ \text{V} − 0.13 V . What is its potential?
Mistakes to avoid Anode minus cathode. Always cathode minus anode. Doubling E ∘ E^\circ E ∘ with the equation. E ∘ E^\circ E ∘ does not change. Calling the anode positive. In a cell that makes electricity it is negative. Electrons in the salt bridge. Only ions move there. Losing the minus sign in Δ G ∘ = − n F E ∘ \Delta G^\circ = -nFE^\circ Δ G ∘ = − n F E ∘ . Check yourself Which electrode is the cathode in a Daniell cell? Stuck? Show a hint Show the answerDoes silver dissolve in a solution of copper sulphate? Stuck? Show a hint Show the answerA cell has E ∘ = 0.50 V E^\circ = 0.50\ \text{V} E ∘ = 0.50 V and n = 2 n = 2 n = 2 . Is Δ G ∘ \Delta G^\circ Δ G ∘ positive or negative? Stuck? Show a hint Show the answerWhy does zinc protect iron from rusting? Stuck? Show a hint Show the answer