Presentation on theme: "Redox Equilibria. Redox equilibria When a metal electrode is placed into a solution of one of its salts two things can happen; 1) Metal ions go into solution;"— Presentation transcript:
Redox equilibria When a metal electrode is placed into a solution of one of its salts two things can happen; 1) Metal ions go into solution; M (s) → M + (aq) + e- 2) metal ions come out of solution. M + (aq) + e- → M (s) The result is a redox equilibrium; M + (aq) + e- ⇌ M (s)
Eg; Cu in a solution of CuSO 4. Copper ions might come out of solution; Cu 2+ (aq) +2e- → Cu (s) Or copper atoms might go into solution; Cu (s) → Cu 2+ (aq) +2e- Giving an eqm; Cu 2+ (aq) +2e- ⇌ Cu (s)
Similar equilbria exist between; 1) Ions of the same metal in different oxidation states with a platinum or graphite electrode. Eg; A mixed solution of iron (ii) and iron (iii) Fe 3+ (aq) +e- ⇌ Fe 2+ (aq) With an inert graphite electrode. NB Platinum is inert and under normal circumstances will not form ions.
H 2 at 101kPa and 298K Platinum wire Porous platinum 2) Ions of a gas in contact with particles of that gas absorbed onto the surface of a platinised platinum electrode. IE; A platinum electrode on which a black, spongy layer of more platinum has been deposited by electrolysis. Acidic solution
The charge on the metal electrode depends on the position of the equilibrium; M + (aq) + e- ⇌ M (s) If the forward reaction is favoured electrons are “removed” so the electrode will have a positive charge. If the backward reaction is favoured; Electrons are “released”, giving a negative charge.
Electrode charges The potential difference between the electrode and solution depends upon; Silver Iron 1) The metal involved 2) The concentration of the solution 0.05M 0.1M
Electrochemical cells Electrochemical cells convert chemical energy into electrical energy. They consist of two half cells, each of which typically has a metal electrode dipping into a salt solution. The half cells are connected by a salt bridge, which allows charge to be transferred between the half cells without allowing the solutions to mix. As current flows electrons will be transferred from the more reactive to the less reactive metal. This is the basic principle behind batteries!
Eg; The Daniel Cell Consists of two half cells; A zinc electrode in a solution of zinc sulphate. A copper electrode in copper sulphate.
The two half cells are then connected via a piece of filter paper soaked in KNO 3 – which avoids the complication of using a metal wire. Salt bridge Cu half cell Zn half cell EMF
Redox reactions in electrochemical cells. Oxidation takes place in the zinc half cell; Reduction takes place in the copper half cell; Zn (s) → Zn 2+ (aq) + 2e-Cu 2+ (aq) +2e- →Cu (s)
When the circuit is completed electrons flow from the more negative to the more positive electrode. Zinc will enter into the solution. Copper will come out of the solution and be deposited on its electrode. Eg in the Daniel Cell from zinc to copper.
Electrode potentials Zn (s) → Zn 2 + (aq) + 2e- The equilibrium will be far to the RHS. Zinc will acquire a –ve charge. It is said to have a –ve electrode potential. Its value is a measure of the ability of the metal to act as a reducing agent (ie to release electrons). Zinc itself will be oxidised.
Potential difference It is not possible to measure the electrode potential of a half cell directly. Only the potential difference (aka voltage) between two half cells. One cell, the hydrogen cell, is used as a reference and allocated a potential of 0. As potential difference depends on temperature and concentration standard conditions are specified. Namely 298 K,101 kPa and a concentration of 1 mol dm -3.
Standard hydrogen electrode H 2 at 101kPa and 298K Platinum wire [H 2 SO 4 ] = 0.5 moldm -3 Porous platinum
Standard Cells One half cell is the standard hydrogen cell… The solution is 1M with respect to the metal ion. … the other is a test cell.
At the standard hydrogen electrode; ½H 2 (g) → H + (aq) + e- The pd is referred to as the standard electrochemical potential, E Θ. For the standard hydrogen electrode E Θ = 0. In the test cell; M + (aq) + e- → M (s) E Θ is the pd measured by the voltmetre in the standard cell (aka electromotive force – emf) Electrodes with –ve E Θ will be better reducing agents than hydrogen. NB as [M + ] increases E Θ becomes more positive as reduction is more likely.
Electrochemical Series Li + (aq) + e- → Li (s) V Ca +2 (aq) + 2e- → Ca (s) V Al +3 (aq) + 3e- → Al (s) V Zn +2 (aq) + 2e- → Zn (s) V Pb +2 (aq) + 2e- → Pb (s) V 2H + (aq) + 2e- → H 2(g) 0.0 V Cu +2 (aq) + 2e- → Cu (s) V Ag + (aq) + e- → Ag (s) +0.8 V The most negative reduction potentials are written at the top of the series. The best oxidising agents at the bottom left. The best reducing agents are at the top right. Redox equilibria are written as reduction.
Cell diagrams Cell diagrams are a shorthand used to represent electrochemical cells. To write one first put down the reactions that are taking place; H 2 → 2H + + 2e-Cu e- → Cu Separate the oxidised and reduced forms by I and leave out the electrons; Pt H 2 I 2H + Cu 2+ I Cu Use a double line, II, for the salt bridge Pt H 2 I 2H + Cu 2+ I Cu II
Caculating the emfs of cells The emf of a cell can be calculated from the E Θ of the half cells. Zn (s) I Zn 2+ (aq) Cu 2+ (aq) I Cu (s) Zn 2+ (aq) + 2e- ⇌ Zn (s) Cu 2+ (aq) + 2e- ⇌ Cu (s) E Θ = VE Θ = +0.34V Look up the E Θ values; EMF = E Θ rhc - E Θ lhc = 0.34 – (-0.76) = 1.1V The cell emf has the sign of the RH electrode – which is the more positive one.
Alternatively instead of using both half equations in the reduced forms, reverse the half cell that involves oxidation; Zn 2+ (aq) + 2e- →Zn (s) E Θ = V Zn (s) → Zn 2+ (aq) + 2e- E Θ = +0.76V Then add the of E Θ the other half cell. Cu 2+ (aq) + 2e- → Cu (s) E Θ = +0.34V EMF = = +1.1V
What would be the emf of a cell formed by connecting up the following half cells? Cu +2 (aq) + 2e- → Cu (s) Ag + (aq) + e- → Ag (s) E Θ = V E Θ = +0.8 V
Cu +2 (aq) + 2e- → Cu (s) Ag + (aq) + e- → Ag (s) Overall EMF = 0.8 – 0.34 = 0.46V E Θ = +0.8 V E Θ = V
What would be the emf of a cell formed by connecting up the following half cells? Cu +2 (aq) + 2e- → Cu (s) Fe 3+ (aq) + e- → Fe 2+ (aq) E Θ = V E Θ = V
Overall EMF = 0.77 – 0.34 = 0.43V Cu +2 (aq) + 2e- → Cu (s) Fe 3+ (aq) + e- → Fe 2+ (aq) E Θ = V E Θ = V
Predicting whether reactions will take place. E Θ s can be used to predict which redox reactions are possible. The more positive E Θ the more likely the reaction. Generally if E Θ ≥ +0.4V a reaction is feasible. But in practice the rate might be so slow that a reaction will not happen. E Θ is defined for standard conditions, if concentrations are increased the reaction may become possible.
Can iodine oxidise bromide ions? 1) Write out the equation; I 2 + 2Br - → 2I - + Br 2 2) Separate this into half equations; I 2 + 2e- → 2I - E Θ = V 2Br - → Br 2 + 2e- E Θ = V 3) Add (-1.09) = -0.55V. 4) < 0.4 so the reaction is not possible.
Competition for electrons E Θ of a half cell is a measure of its oxidising or reducing power. IE its ability to compete for electrons. Generally stronger oxidising agents have the more positive E Θ s. Eg; I 2 + 2e- → 2I - E Θ = V Cl 2 + 2e- → 2Cl - E Θ = V Therefore chlorine is a stronger oxidising agent than iodine.
The more negative E Θ s the stronger is the reducing agent. Eg; Pb e- ⇌ Pb E Θ = -0.13V Ca e- ⇌ Ca E Θ = -2.87V Therefore calcium is a stronger reducing agent than lead.