Electrochemistry Flashcards
in terms of electrons, oxidation is
the loss of electrons
in terms of electrons, reduction is
the gain of electrons
in terms of change in oxidation number, oxidation is
when the oxidation number increases
in terms of change in oxidation number, reduction is
when the oxidation number decreases
the reactions involved in the measurements of a standard electrode potential are……………..reactions in a state of……………….
redox reactions in equilibria
when a metal such as Magnesium or Copper, represented by ‘M’, is placed in water, there is a very small tendency for the metal atoms to…..
lose electrons and go into solution as positive metal ions
write the equations that represents the loss of electrons from either Magnesium or Copper atoms (‘M’ and ‘x’ for the charge) to form aqueous metal ions and electrons: (write the non-reversible equation and the reversible one once equilibrium has been reached)
M(s) → xe- + Mx+(aq)Mx+(aq) + xe- → M(s)xe- + Mx+(aq) ⇌ M(s)
explain the following equations in relation to the establishment of an equilibrium:M(s) → xe- + Mx+(aq)Mx+(aq) + xe- → M(s)xe- + Mx+(aq) ⇌ M(s)
M(s) → xe- + Mx+(aq)-The metal atoms lose electrons and go into solution as positive metal ions, with the electrons remaining on the metal’s surface.-the electrons build up on the metal’s surface and the resulting negative charge attracts the positive metal ions in solution, creating a layer of positive ions surrounding the metalMx+(aq) + xe- → M(s)-some of the metal onis will regain their electrons from the surface of the metal and return to form part of the metalxe- + Mx+(aq) ⇌ M(s)-eventually, a dynamic equilibrium is reached as the rate at which metal atoms leave the metal surface is the same as the rate at which metal ions join the metal from solution
-the equations showing the tendencies of magnesium (Mg) and copper (Cu) to release electrons to form positive ions are:-also show the position of equilibrium for each!
Mg2+(aq) + 2e- ⇌ Mg(s)Mg’s position of equilibrium lies further to the left than for copper, because of magnesium’s greater tendency to release electrons than copper. this means magnesium will have a greater negative charge on the metal surface and more positive ions in solutionCu2+(aq) + 2e- ⇌ Cu(s)
<p>Define standard electrode potential </p>
<p>the potential difference between the metal and the solution</p>
the standard hydrogen electrode consists of
hydrogen gas at a pressure of 100 kPa (1 bar) bubbling over a piece of platinum foil dipped into a solution of hydrochloric acid (or sulfuric acid) with a hydrogen ion concentration of 1 ol dm-3 at a temperature of 298 K
in the standard hydrogen electrode, the platinum foil is covered in ………………………. which allows an equilibrium between the hydrogen ions in solution and the hydrogen gas to be established quickly because……………….
covered in porous platinumporous platinum has a large surface area so allows the equilibrium to be reached more quickly
the standard hydrogen electrode equilibrium equation is
H+(aq) + e- ⇌ 1/2H2(g)
when writing equilibria for standard electrode potentials, we ALWAYS write the electrons on the ………….. hand side
LEFT hand side
the standard conditions that apply to all equilibria so that a fair comparison can be made between them are:
gas pressure, 100 kPatemperature, 298Kconcentration of ions in solution, 1 mol dm-3
describe how you would measure the standard electrode potential of a metal ion | metal system such as Mg2+ | Mg
-set up a standard hydrogen electrode by bubbling H2 gas over a piece of platinum foil at 100 kPa pressure and 298 K-use a solution of HCl acid at a 1 mol dm-3 H+ ion concentration and use porous platinum to cover the platinum foil-clean a piece of magnesium metal with sand paper and attach attach to a connecting wire. also attach a connecting wire to the platinum foil and link both up to a high resistance voltmeter-dip the Mg electrode into a 1 mol dm-3 concentration magnesium sulfate solution-link up both beakers with a salt bridge made of filter paper saturated with potassium nitrate (KNO3)
why is a high resistance voltmeter used?
ideally this would have infinite resistance so there would be no flow of electrons (no current) around the external circuit so that the reading would represent the difference in potential between the two half cells when both reactions are at equilibrium.
why is an ionic salt used in the salt bridge, and why is it usually KNO3 ?
the ionic salt allows the movement of ions between the half-cells and its usually KNO3 because the salt bridge ionic salt cannot interact with any of the half-cell ions, and usually KNO3 is suitable
<p>E⦵cell =</p>
<p>E⦵red - E⦵ox</p>
the standard electrode potential, E⦵ , of the standard hydrogen electrode is
E⦵ = 0.00 V
the E⦵ value for Mg2+(aq) | Mg(s) is -2.37V while the value for Cu2+(aq) | Cu(s) is +0.34V. what do the signs on the E⦵ values tell us about the relative reducing and power of magnesium and copper
the negative value for Mg2+(aq) | Mg(s) tells us the equilibrium position if further tot he left than the equilibrium position. the positive sign for Cu2+(aq) | Cu(s) indicates the position of equilibrium of this reaction of further to the right than the equilibrium positions of the reaction in the hydrogen electrode.therefore, magnesium releases electrons more readily than hydrogen and that copper releases electrons less readily than hydrogen.this means magnesium is a better reducing agent than both hydrogen and copper, while hydrogen is better reducing agent than copper
a reducing agent is
a species that reduces another species by adding on or more electrons to it, so is therefore itself oxidised
a negative E⦵ means that
the position of equilibrium of the half cell reaction lies FURTHER TO THE LEFT than the equilibrium of the standard hydrogen electrode
a positive E⦵ means
the position of equilibrium of the half cell reaction lies FURTHER TO THE RIGHT than the equilibrium of the standard hydrogen electrode
