Topic 14 - Further Redox Flashcards
What is meant by the term standard electrode (redox) potential?
Reactions involved in measurement of standard electrode potential are redox reactions. The reactions involved are equilibria.
Explain the different tendencies of metals to release electrons to form positive ions?
Background…
Equations/Examples? Metal in water? (Mg and Cu tendencies, position of equilibrium ect.)
Look at the term absolute potential difference with Mg and Cu
When metal like Mg or Cu placed in water, small tendency for metal atoms to lose electrons and go in solution as positive ions…
M(s) —> M(n+)(aq) + ne- (n = any number)
Electrons remain on the surface of the metal. In a short time there is a build up of electrons on surface of metal which results in a - charge attracting + ions. Layers of + ions formed surrounding the metal.
Some of the + ions regain their electrons from the surface of the metal and return to form part of the metal surface… M(n+)(aq) + ne- —> M(s)
Dynamic equilibrium will be established in which rate a which ions leaving the surface of the metal to go into solution is the same as the rate at which they are joining it from solution
If look at Mg and Cu the difference between Mg and Cu is that the equilibrium position will be further to the left for Mg because tendency of Mg to release electrons is greater, greater - charge on the metal and more + ions in solution. The pd is greater with Mg - absolute pd is measure the pd between metal and the solution. But not possible to measure it between metal electrode and solution as there would be an extra piece of metal in solution that could connect the solution to a voltmeter
Explain the features of a standard hydrogen electrode and explain why a reference electrode is necessary
The reference electrode of choice is the standard hydrogen electrode.
The electrode consists of H gas at pressure of 100kPa bubbled over piece of platinum foil dipped in HCl with H+ conc of 1M at 298K.
Surface of foil is porous platinum which has a large surface area and allows an equilibrium between H+ and H gas to be established.
Explain process of measuring standard electrode potential… look at E values, features of electrode, process/steps, example between magnesium and magnesium ions
Connect standard hydrogen electrode to magnesium electrode via circuit with high resistance voltmeter. Salt bridge used containing conc solution of potassium nitrate which allows the movement of ions. If the right hand cell contains silver nitrate then potassium chloride shouldnt be used as ppte would form.
High resistance voltmeter used as voltmeter should have infinite resistance so there is no flow of electrons around external circuit, if possible then reading on voltmeter would represent difference in potential between 2 hald cells where both reactions in equilibrium
With Mg connected to hydrogen electrode, reading is 2.37V. If Mg replaced by copper half cell, reading would be 0.34V. If meter was taken out of circuit, electrons would flow from Mg electrode to hydrogen electrode so Mg the negative electrode. Copper, electrons from hydrogen to copper electrode making copper the positive. DIfference in potential measured called the standard electrode potential.
What do E values tell us?
The more negative the E value, the further the equilibrium lies to the left, more readily the metal loses electrons to form ions while the more positive the E value, the further the equilibrium lies to the right, less readily, the metal loses electrons to form ions
What is electromotive force?
When we determine a standard electrode potential, the pd measured when no electrons are flowing through the external circuit. Pd measured is called electromotive force (emf) of the cell… standard emf given symbol Ecell. Can be + or - depending ob reference point against the pd is measured. Reference point of standard hydrogen electrode is set at 0. Emf of a cell in which hydrogen electrode is + will have negative value.
Explain measuring standard electrode potentials of more complicated redox system
1) Systems involving gases
2) Non metal elements and their ions in solution
3) ions of the same element with different oxidation numbers
1) Measuring Cl2 gas to Chloride ions
Setting up a half cell which chlorine gas bubbled into solution with chloride ions. To established equilibrium, need an electrical connection to external circuit so would need a piece of platinum in solution. Half cell connected to standard hydrogen electroe and emf measured in usual way… measured at 1.36V
2) Examples could be Bromine to Bromide or Iodine to Iodide. To determine Ecell, half cell contains solution of Br2 and Bromide ions with conc of 1M connected to standard hydrogen electrode… Similar set up for Iodine too.
3) Ions of same element could be Fe3+ and Fe2+… need platinum wire and platinum foil covered in porous platinum, increase surface area. Conc of each ion is 1M
Determining the best reducing and oxidising agents from E values
Species on right hand side of half cell reactions are capable of behaving as reducing agents because they can lose electrons. Most powerful reducing agent is lithium becaue its redox system is the most - E value equilibrium position is furtherest to the left.
Leas powerful has least negative value so position of eq. Is furtherest to the right. Species on left side of half cell reactions capable of acting as oxidising agents - least negative value strongest oxidising agent
What is an electrochemical cell? Look at example with Zinc and Copper Determine the overall cell reaction and the cell diagram from the data... Zn2+ + 2e- ----> Zn(s) = -0.76V Cu2+ + 2e- ----> Cu(s) = +0.34V
Devide for producing an electric current from chemical reactions. Constructed from two half cells.
One constructed from Zn2+ | Zn half cell and Cu2+ | Cu half cell under standard conditions
E value for Zn more negative so zinc electrode will be the negastive electrode… electrons will flow through the external circuit from zinc to copper
Half cell notations used to show the two reduced forms of the species shown on outside of cell diagram, postivie electron shown on right hand side, solid line is phase boundary. Double vertical line indicates a salt bridge.
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
-0.76V. +0.34V
Difference is 1.10 so emf is +1.10V right - left
What is meant by the difference between two standard electrode potentials?
Calculate the emf
Making predictions using standard electrode potentials to measure how easily a species loses electrons?
Method provides info on how far to the left an eq is relative to the eq in the standard hydrogen electrode. For example, zinc lies further to the left than the hydrogen electrode and copper lies further to the right than the hydrogen electrode. If combined, the electrons flow from the zinc to copper electrode so electrons flow from half cell with more negative half cell to a more positive E value.
Making predictions for thermodynamically feasibility of the reaction…
Zn(s) + Cu2+(aq) —-> Zn2+(aq) + Cu(s)
Will zinc and copper react with dilute sulfuric acid?
Look at E values for the half cell reactions…
Zn2+(aq) + 2e- —-> Zn(s) = -0.76V
Cu2+(aq) + 2e- —-> Cu(s) = +0.34V
E value for zinc more negative so position of eq will shift to the left releasing electrons and position of Cu eq will shift to the right, accepting electrons so the reaction is thermodynamically feasible
As the zinc electrode is more negative than hydrogen, the reaction is thermodynamically feasible because zinc will release electrons to hydrogen ions
Copper is more positive than hydrogen so reaction not thermodynamically feasible because copper will not release electrons to hydrogen ions, reverse reaction is thermodynamically feasible but no reaction takes place when H is bubbled into aqueous solution with copper ions as the activation energy from reaction is very large so reactants are kinetically stable
Describe the reaction between manganese(IV) oxide and hydrochloric acid
Mention E values, thermodynamic feasibility
Preparing chlorine in the lab, react these two togther.
Standard electrode potentials suggest not feasible as Manganese at +1.23V and Chlorine at +1.36V
Equilibrium 2 more positive so chloride ions cannot release electrons to MnO2… required for reaction to take place. Reaction not thermodynamically feasible under standard conditions in which conc of HCl and conc of both H and Cl ions are 1M
Increase conc of Hcl - if concentrated used, concentration of ions increase so shifts position of equilibrium 1 to the right and 2 to the left
As a result electrode potnetial for eq 1 becomes more positibe beaude redox system better electron acceptors.
Electrode potential of eq 2 becomes less positive than eq1 so chloride ions can now release electrons to maganese(IV) oxide so thermodynamically feasible under nonstandard conditions
Understand how disproportionation reactions can relate to standard electrode potentials
Typical one is conversion of copper(I) ions in aq solution to copper(II) ions and copper atoms
Cu2+(aq) + e- —-> Cu+(aq). +0.15V
Cu+(aq) + e- —–> Cu(s) +0.52V
E value for eq 1 is more negative than 2 so position will move to the left and eq 2 will move to the right so Cu+ ions will both release and accept electrons to one another to form Cu2+ ions and Cu atoms - Cu+ ions disproportionate
Describe relationship between total entropy and Ecell
Standard Gibbs energy change taking place in a cell and the emf of the cell related by expression…
Gibbs = -nFEcell
n = number of moles of electrons and F is the Faraday constant
Gibbs = DeltaH - T x DeltaSsystem
So T x DeltaStotal = nFEcell
N is a constant for given cell reaction and F is a constant so follows that at given temp, the total entropy change of the cell reaction proportional to the emf of the cell
If Ecell is positive then reaction as written from left to right in diagram is thermodynamically feasile because DeltaStotal would be positive
