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Flashcards

Topic 14

Jade B Chemistry GCSE English 72 cards

Topic 14

Cards in this deck (72)

  1. 1The electrode potential of a ........................................ in contact with its aqueous ................ can be measured (fill in the blanks)metal or non-metal
    ions
  2. 2Define reduction (3 ways)- loss of oxygen
    -gain of electrons
    -decrease in oxidation number
  3. 3Define oxidation (3 ways)- gain of oxygen
    -loss of electrons
    -increase in oxidation number
  4. 4what is a redox reactiona reaction which involves both reduction and oxidation
  5. 5what happens when manganate (VII) ions react with iron (II) ions Give the chemical equationManaganate (VII) ions oxidise iron (II) ions
    Mn04 - + 8h+ + 5fe 2+ = Mn 2+ + 4h20 + 5fe 3 +
    - manganate (VII) ions recuced to manganese(II)
    - Iron (II) oxidised to iron (III)
  6. 6from the above reaction which is acting as the oxidising agent?- Manganate (VII) is oxidising agent
    -Iron (III) is reducing agent
  7. 7What is a disproportionation reaction?- when an element in a single species is simultaneously oxidised + reduced
  8. 8explain the disproportionation reaction occurring below (including oxidation states) CU20 (s) + H2SO4 (aq) = Cu (s) + CuSo4 (aq) + H20 (l)copper (I) is reduced to copper (+1 .......... 0)
    Copper (I) is oxidised to copper (II) (+1...........+2)
  9. 9what happens when a metal is placed in a solution of its ions? What does it form?- an equilibrium establishes
    - An electrical potential (E) exists between metal + solution causing a half- cell to form
  10. 10how can you measure the electrode potential of a half cell?- must be measured against a reference electrode (e.g. standard hydrogen electrode)
    - cant be measured directly
  11. 11What is a standard electrode potential (E bus stop sign) of a half cell?- the electromotive force (EMF) of a cell
    - where the standard hydrogen electrode is on the left + the half cell is on the right
    - at 298K, 100kPA (pressure of gases involved), 1.00moldm-3 conc of ions
  12. 12what is emf?- the potential difference (voltage\0 measured using a high-resistance voltmeter
  13. 13Why is a platinum electrode used in the standard hydrogen electrode?- Platinum is unreactive
    -present as a porous platinum (holey) enabling
    - large S.A for reaction
    -allows conduction of electrons
  14. 14Draw a diagram of the standard hydrogen electrode (with labels)Draw a diagram of the standard hydrogen electrode (with labels)
  15. 15How do you calculate E values? How do you calculate E cell values?- standard hydrogen electrode used to measure E values
    - E cell values found from the E values for 2 half cells
  16. 16How do you get a electrochemical cell?-by combing 2 half-cells
  17. 17what connections are needed between a cell?- an external electrical circuit : connects the 2 electrodes + allows electrons to flow from 1 half cell to the other
    - a salt bridge : connects the 2 solutions + allows ions to move between 2 half-cells
  18. 18give a common salt bridge used why is this salt bridge often used?- strip of filtered paper soaked in concentrated potassium nitrate solution
    - avoid the possibility of a precipitate forming with ions from half-cells
  19. 19how are half-cells represented? Give this for Mg (s), Mg 2+ (aq)- using cell diagrams
    - Mg 2+ (aq) / Mg (s)
  20. 20what does the vertical line represent in cell diagrams? How do you separate species in the same phase (state) ?- a phase boundary
    - via a comma
  21. 21draw the cell diagam for a standard hydrogen electdoe + magnesium 2+ half cellH2 (g) / H+ (aq) : : Mg 2+ (aq) / Mg (s)
    -Oxidation goes on left and reduction on right
    - dotted line in middle= salt bridge
  22. 22Give the equation for emf under standard conditions Give the standard electrode potential for the standard hydrogen electrode- E cell = E right - E left
    - for hydrogen electrode = 0V
  23. 23What does E stand for? What does E cell stand for?- E= Standard electrode potential
    -E cell = EMF
  24. 24how do we measure the EMF value for a cell?- construct 2 half cells
    - connecting them via a salt bridge + external circuit
    - maintaining standard conditions
    - measuring the potential by a high-resistance voltmeter
  25. 25Sketch the set up of a cell diagram for Zn and CuSketch the set up of a cell diagram for Zn and Cu
  26. 26—
  27. 27what does the electrode potentials tell us about oxidising and reducing agents?- the greater the negative electrode potential the more powerful the reducing agent
    - the greater the positive electrode potential the more powerful the oxidising agent
  28. 28Fe 3+ (aq) + E- = Fe2+ (aq) e value is +0.77 What does the e value tell you?- that Iron (III) ions is a powerful oxidising agent
    -When value positive the agent Is one on left side
  29. 29Mg2+ (aq) + 2E- = Mg (s) e value is -2.37 What does this tell you?that magnesium is the most powerful reducing agent
    - when negative the agent is on the right side
  30. 30How do you predict the feasibility of a reaction- Feasibly if e cell value is positive
    - write E value for 2 electrode systems (with the more positive E value on right)
    - the reduction will be on the right, so write as seen in table
    - oxidation will be on left so write half equation in reverse
    - combine 2 half equations
  31. 31Is this reaction feasible? 1) Cu2+ (aq) + e- = Cu+ (aq) 2) Cu + (aq) + e- = Cu (s)1)E= +0.15V
    2) E= +0.52V
    As no.2 more positive it is found on right
    e.g. E cell= E right - E left
    = +0.52-0.15
    = +0.37 V = positive= feasible
  32. 32Using 1 + 2 write its full equation What type of reaction is this2Cu + (aq) = Cu (s) + Cu2+ (aq)
    1) Half equation written forward as most positive
    2) written backwards as most negative
    Reaction is a disproportionation for cu+ releases electrons to form cu2+ and gains them to form Cu atoms
  33. 33will all reactions that are thermodynamically feasible work ?In practice not all reactions will work
  34. 34how can you predict if a reaction is thermodynamically feasible ?- by using the E values
    e.g. E cell = E right - E left
  35. 35give 2 reasons why a reaction may not occur- the activation energy (Ea) may be very large
    -the conditions for the reaction may not be standard
  36. 36if the conditions aren't standard will a thermodynamic feasible reaction never happen?- Yes
    - But a non-feasible reaction may if conditions are altered
    - the position of equilibrium of a half cell may change = altered E
    - which changes the Ecell which may become positive as a result
  37. 37How do you calculate Gibbs energy G using E cell values?G= -nFEcell
    n= moles of electrons
    F= Faraday constant
  38. 38how does entropy change in a system (S) relate to emf of a cell at a given temperature?E cell = Entropy change
    Standard emf of a cell is directly proportional to the total entropy change
  39. 39Does S and E cell need to both be positive for a cell reaction to be feasible?-Yes
    -as if E cell is positive = S is positive
  40. 40What changes the position of an equilibrium in a half cell?- if the concentration of a reactant is changed
  41. 41Mg2+ (aq) + 2e- = Mg (s) E cell = -2.37V what happens if the conc of Mg2+ is decreased?- position of equilibrium moves to the left
    - more electrons are released
    - electrode become more negative
    -E become more negative (half cell)
    - When is found on left hand (e.g. when reacting with Cu, a positive value) the E cell increases
  42. 42Using E cell, R,T, nF, InK put them into an equasionE cell = RT/nF x InK
  43. 43R,n and F are constants so at a given temp E cell = InK Explain what this means-at a given temp the standard emf of a cell is directly proportional to the natural logarithm of the equilibrium constant
    - meaning if K is greater than one then E cell is positive
  44. 44—-Copper(II) ions are more concentrated in right hand cell.
    -Copper(II) ions + electrons will be released in the left hand cell in preference to right hand
    - so left hand electrode more negative    -Copper(II) ions are more concentrated in right hand cell. -Copper(II) ions + electrons will be released in the left hand cell in prefere...
  45. 45how does current flow affect the potential difference? When the current is zero what does it equal?- when a current flows the measured potenetial difference decreases
    - when 0 the reading = to Ecell
  46. 46why do we used a high-resistance voltmeter in a cell?- allow negligible current to flow
  47. 47What are fuel cells?cells that generate a voltage by the reaction of hydrogen, methanol or other hydrogen- rich fuels with oxygen
  48. 481 type of fuel cells is the hydrogen-oxygen fuel cell give the reaction that occurs within it.H2 (g) + 1/2 02(g) = H20(l)
  49. 49Explain hoe hydrogen-oxygen fuel cells work- produce electricity directly from the reaction between hydrogen + oxygen (from air)
    - will produce electricity as long as reactants are supplied
    - makes water vapour as only product
  50. 50give 1 advantage + disadvantage of the use of hydrogen-oxygen fuel cellspositive- don't release Co2 (greenhouse gas)
    Negative- needs hydrogen, which is explosive + difficult to store
  51. 51Give another use of hydrogen beyond as a fuel in fuel cellsused in internal combustion engines
  52. 52draw a diagram of a hydrogen oxygen fuel cell with an acidic electrolyte- fully labeled (components of a fuel cell)draw a diagram of a hydrogen oxygen fuel cell with an acidic electrolyte- fully labeled (components of a fuel cell)
  53. 53in the previous question the fuel cells contains electrodes what is special about them?- the electrodes are coated with platinum
  54. 54Give the reaction at the negative electrode of a hydrogen-oxygen fuel cell with an acidic electrolyteH2(g) = 2H+ (aq) + 2e-
  55. 55Give the reaction at the positive electrode of a hydrogen-oxygen fuel cell with an acidic electrolyte1/2 O2(g) + 2H+ (aq) + 2e- = H20 (l)
  56. 56explain what happens to the hydrogen ins in the fuel cells that contains an acidic electrolyte- hydrogen ions formed at negative electrode
    - than pass through a proton exchange membrane to the positive electrode
  57. 57give the overall reaction that occurs in a hydrogen-oxygen fuel cell (regardless of If it contains acidic electrolyte or alkaline one)H2(g) + 1/2 02 (g) = H20 (l)
  58. 58Give the reaction at the negative electrode of a hydrogen-oxygen fuel cell with an alkaline electrolyteH2(g) + 2OH- (aq) = 2H20 (l) + 2e-
  59. 59Give the reaction at the positive electrode of a hydrogen-oxygen fuel cell with an alkaline electrolyte1/2 02 (g) + H20 (l) + 2e- = 2OH-(aq)
  60. 60explain what happens to the water molecules in the fuel cells that contains an alkaline electrolyte. What about hydroxide ions?-water is formed at the negative electrode
    -it then diffuses through membrane to the positive electode
    - hydroxide ions are formed at the positive eectrode
    - duffuse throught membrane to negative electrode
  61. 61what are storage cells?-rechargeable batteries
    - they can be recharged because their cell reactions can be reversed by applying a voltage
  62. 62give examples of storage cells- lithium ion cells
    -nickel-cadmium cells
    -lead-acid batteries
  63. 63does oxidation or reduction occur at the negative electrode? the positive electrodeat negative electrode electrons are lost (oxidation) e.g. Li= Li+ + e-
    at positive electrode electrons are added (reduction) e.g Li+ + Mn02 + e- = LiMnO2
  64. 64A methanol-oxygen fuel cell uses methanol as its fuel. When in use methanol reacts with water to from carbon dioxide + hydrogen ions Write a half-equation for the reaction that occurs as the methanol (negative) electrodeCH30H + H20 = CO2 + 6H+ + 6E-
  65. 65Still in previous cards set up Write a half-equation for the reaction that occurs as the positive electrode1/202 + 2h+ + 2e- = H20
  66. 66If have time go through worked examples on pg99 + 100 in revision guide....
  67. 67what are redox titrations used to determine?the conc of a reactant
    e.g. iron(II) ions or iodide
  68. 68Manganate (VII) ions can be used in redox titrations to determine the conc of iron(II) ions, explain howComplete a titration
    1) Iron (II) solution is acidified with dilute H2SO4 (provides H+ (aq))
    2) KMnO4 (aq) is added from a burette
    3) The KMnO4 (aq) turns from colourless in the solution until the end point, than 1 more drop turns the mixture pale pink
  69. 69MnO4- and Mn2+ have which colours associated with them?Mno4- = purple
    Mn2+ = pale pink
  70. 70what is used in a redox titration to determine the conc of an iodine solution? Explain this process- Thiosulfate ions
    1) iodine solution is pipetted into a flask
    2) Na2SO3 2- (thiosulfate ions) added from a burette
    3) iodine solution turns brown to pale yellow near the end point
    4) a few drops of starch indicator are added
    5) mixture than turns from blue to colourless at the end point
  71. 71iron (II) sulfate tablets are sold as a dietary supplement. One 0.5g tablet was dissolve in dilute H2SO4. The titre was 22.40cm3 when the solution was titrated against 0.010 moldm-3 KMnO4 (aq). Calculate the % of iron in the tabletAmount of MnO4- = 0.010 x 22.40 x 10-3
    =2.24 x 10-4 mol
    Amount of Fe2+ = 5 x 2.24 x 10-4
    =1.12 x 10-3 mol
    Mass of Fe = 1.12 x 10-3 x 55.8
    =0.0625 g
    % of iron = 100 x 0.0625 / 0.5
    =12.5%
  72. 72go through worked example pg 101 6 mark question

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GCSE Chemistry (Topic 10-14 ) Flashcards on Topic 14, created by Jade B on 04/02/2018.
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