Flashcards
Topic 14
Flashcards do conjunto (72)
- 1The electrode potential of a ........................................ in contact with its aqueous ................ can be measured (fill in the blanks)metal or non-metal
ions - 2Define reduction (3 ways)- loss of oxygen
-gain of electrons
-decrease in oxidation number - 3Define oxidation (3 ways)- gain of oxygen
-loss of electrons
-increase in oxidation number - 4what is a redox reactiona reaction which involves both reduction and oxidation
- 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) - 6from the above reaction which is acting as the oxidising agent?- Manganate (VII) is oxidising agent
-Iron (III) is reducing agent - 7What is a disproportionation reaction?- when an element in a single species is simultaneously oxidised + reduced
- 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) - 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 - 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 - 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 - 12what is emf?- the potential difference (voltage\0 measured using a high-resistance voltmeter
- 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 - 14Draw a diagram of the standard hydrogen electrode (with labels)
- 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 - 16How do you get a electrochemical cell?-by combing 2 half-cells
- 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 - 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 - 19how are half-cells represented? Give this for Mg (s), Mg 2+ (aq)- using cell diagrams
- Mg 2+ (aq) / Mg (s) - 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 - 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 - 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 - 23What does E stand for? What does E cell stand for?- E= Standard electrode potential
-E cell = EMF - 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 - 25Sketch the set up of a cell diagram for Zn and Cu
- 26—
- 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 - 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 - 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 - 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 - 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 - 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 - 33will all reactions that are thermodynamically feasible work ?In practice not all reactions will work
- 34how can you predict if a reaction is thermodynamically feasible ?- by using the E values
e.g. E cell = E right - E left - 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 - 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 - 37How do you calculate Gibbs energy G using E cell values?G= -nFEcell
n= moles of electrons
F= Faraday constant - 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 - 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 - 40What changes the position of an equilibrium in a half cell?- if the concentration of a reactant is changed
- 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 - 42Using E cell, R,T, nF, InK put them into an equasionE cell = RT/nF x InK
- 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—-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
- 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 - 46why do we used a high-resistance voltmeter in a cell?- allow negligible current to flow
- 47What are fuel cells?cells that generate a voltage by the reaction of hydrogen, methanol or other hydrogen- rich fuels with oxygen
- 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)
- 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 - 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 - 51Give another use of hydrogen beyond as a fuel in fuel cellsused in internal combustion engines
- 52draw a diagram of a hydrogen oxygen fuel cell with an acidic electrolyte- fully labeled (components of a fuel cell)
- 53in the previous question the fuel cells contains electrodes what is special about them?- the electrodes are coated with platinum
- 54Give the reaction at the negative electrode of a hydrogen-oxygen fuel cell with an acidic electrolyteH2(g) = 2H+ (aq) + 2e-
- 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)
- 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 - 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)
- 58Give the reaction at the negative electrode of a hydrogen-oxygen fuel cell with an alkaline electrolyteH2(g) + 2OH- (aq) = 2H20 (l) + 2e-
- 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)
- 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 - 61what are storage cells?-rechargeable batteries
- they can be recharged because their cell reactions can be reversed by applying a voltage - 62give examples of storage cells- lithium ion cells
-nickel-cadmium cells
-lead-acid batteries - 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 - 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-
- 65Still in previous cards set up Write a half-equation for the reaction that occurs as the positive electrode1/202 + 2h+ + 2e- = H20
- 66If have time go through worked examples on pg99 + 100 in revision guide....
- 67what are redox titrations used to determine?the conc of a reactant
e.g. iron(II) ions or iodide - 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 - 69MnO4- and Mn2+ have which colours associated with them?Mno4- = purple
Mn2+ = pale pink - 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 - 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% - 72go through worked example pg 101 6 mark question
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- GCSE Chemistry (Topic 10-14 ) FlashCards sobre Topic 14, criado por Jade B em 04-02-2018.
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