2-July-2016 Chemsheets A Page 1

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1 2-July-2016 Chemsheets A Page 1

2 SECTION 1 AS REDOX REVISION 1) Oxidatin states When using xidatin states, we effectively imagine everything t be an in the xidatin state is the charge it wuld have if it was an in. Assigning xidatin states: 1) On simple ins, the xidatin state is the charge n the in e.g. Cu 2+ (Cu +2); Cl (Cl 1); Al 2O 3 (Al +3, O -2) 2) In elements, the xidatin state is always zer e.g. Cl 2 (Cl 0) 3) The ttal f all the xidatin states must always equal the verall charge n the species. 4) In mlecules and mre cmplex ins, the mre electrnegative element is assumed t be the negative in 5) H is nearly always +1 and xygen 2 e.g. CH 4 (C -4, H +1); CO 2 (C +4, O -2); H 2O (H +1, O -2) TASK 1 Oxidatin states 1) Calculate the xidatin state f the stated element in the fllwing species: a) Fe in Fe FeCl 3 FeCl 2 K 2FeO 4 [Fe(H 2O) 6] 2+ b) Cl in Cl 2 ClO 3 - ClO Cl 2O 7 NaCl 2) Calculate the xidatin state f each element in the fllwing: SO 2 S O H 2O 2 H O S S NaH Na H SO 3 S O 2 CO 3 C O H 2S H S Cr 2O 3 Cr O NH 3 N H CrO 3 Cr O NO 2 N O KMnO 4 K Mn O N 2 N K 2MnO 4 K Mn O NO 3 N O Cu 2O Cu O Cl Cl CuO Cu O SO 4 2 S O NaCuCl 2 Na Cu Cl 3) Give the xidatin state f the transitin metal in each f the fllwing species. a) [C(H 2O) 6] 2+ C e) [FeO 4] 2 Fe i) K 2[CCl 4] C b) [CrCl 6] 3 Cr f) [Mn(CN) 6] 4 Mn j) K 3[AuF 6] Au c) [C(NH 3) 6]Cl 2 C g) [Ni(CO) 4] Ni k) (NH 4) 2[IrCl 6] Ir d) [C(NH 3) 5Cl]Cl 2 C h) [Ni(EDTA)] 2 Ni l) Na[Mn(CO) 5] Mn 2-July-2016 Chemsheets A Page 2

3 2) Writing half equatins MnO 4 Mn 2+ N 2O NO 3 1) Wrk ut xidatin states Mn(+7) t Mn(+2) N(+1) t N(+5) 2) Balance the element which changes xidatin state already balanced There are 2N n the left s we need 2NO 3 n the right t balance the Ns N 2O 2NO 3 3) Balance the electrns Mn(+7) gains 5e t frm Mn(+2) MnO e Mn 2+ 4) Balance O s with H 2O Need t add 4O s n the right s add 4H 2O n the right MnO 4 + 5e Mn H 2O 5) Balance H s with H + Need t add 8H s n the left s add 8H + n the left MnO 4 + 5e + 8H + Mn H 2O N(+1) lses 4e t frm N(+5); as there are 2N that each lse 4e, then 8e are lst N 2O 2NO e Need t add 5O s n the left s add 5H 2O n the left N 2O + 5H 2O 2NO e Need t add 10H s n the right s add 10H + n the right N 2O + 5H 2O 2NO 3 + 8e + 10H + 6) Check half equatin by checking ttal charge n bth sides f the equatin is the same (if they are the same then the equatin is prbably right if they are nt the same it is definitely wrng!) LHS ttal charge = = 2+ RHS ttal charge = 2+ LHS ttal charge = 0 RHS ttal charge = = 0 TASK 2 Writing half equatins Write half equatins fr each f the fllwing cnversins. a) I I 2... b) Fe 2+ Fe c) Fe Fe d) Cl Cl 2... e) SO 4 2- SO 2... f) VO 2+ VO g) SO 4 2 H 2S... h) H 2O 2 O 2... i) Cr 3+ Cr 2O j) C 2O 4 2 CO 2... k) Hg 2 2+ Hg... l) lo 3 l July-2016 Chemsheets A Page 3

4 3) Cmbining half equatins e.g. cmbine these tw half equatins: MnO 4 + 5e + 8H + Mn H 2O (equatin 1) Fe 2+ Fe 3+ + e (equatin 2) Step 1 balance the electrns: Equatin 2 x5 t balance the electrns Equatin 1 MnO 4 + 5e + 8H + Mn H 2O Equatin 2 x5 5Fe 2+ 5Fe e Step 2 add the half equatins tgether (the electrns shuld cancel): MnO 4 + 5e + 8H + + 5Fe 2+ Mn H 2O + 5Fe e Final equatin: MnO 4 + 8H + + 5Fe 2+ Mn H 2O + 5Fe 3+ TASK 3 Cmbining half equatins Cmbine these half equatins t prduce redx reactins. 1) Zn Zn e Fe 3+ + e Fe 2+ 2) 2 Cr 2O 7 + 6e + 14H + 2Cr H 2O Fe 2+ Fe 3+ + e 3) H 2SO 4 + 8e + 8H + H 2S + 4H 2O 2I I 2 + 2e 4) MnO 4 + 5e + 8H + Mn H 2O 2Cl Cl 2 + 2e 5) MnO 4 + 5e + 8H + Mn H 2O H 2O 2 O 2 + 2H + + 2e 6) 2I I 2 + 2e 2IO e + 12H + I 2 + 6H 2O 2-July-2016 Chemsheets A Page 4

5 4) Redx prcesses O xidatin I s L ss R eductin I s G ain f electrns A redx reactin is ne in which ne r mre elements change xidatin state. e.g. CuSO 4 + Zn Cu + ZnSO 4 Cu +2 0 S O -2-2 Zn 0 +2 Cu reduced frm +2 t 0; Zn xidised frm 0 t +2 redx reactin A disprprtinatin reactin is ne in which an element is bth xidised and reduced. e.g. Cl 2 + H 2O HCl + HOCl Cl H Cl reduced frm 0 t -1 and xidised frm 0 t +1 redx reactin O -2-2 (disprprtinatin) An xidising agent (xidant) is the substance that des the xidising (i.e. takes away the electrns frm the substance that is xidised, and s the xidising agent is itself reduced). An reducing agent (reductant) is the substance that des the reducing (i.e. dnates the electrns t the substance that is reduced, and s the reducing agent is itself xidised) TASK 4 Redx reactins r nt? Equatin Redx reactin? Disprprtinati n reactin? Species xidised Species reduced Oxidising agent Reducing agent Fe 2O 3 + 3CO 2Fe + 3CO 2 Mg + 2HCl MgCl 2 + H 2 MgO + 2HCl MgCl 2 + H 2O Al 2O 3 + 2Fe 2Al + Fe 2O 3 [C(H 2O) 6] Cl [CCl 4] H 2O Na 2O + H 2O 2NaOH 2H 2O 2 2H 2O + O 2 2NaBr + H 2SO 4 Na 2SO 4 + 2HBr Cl 2 + 2NaOH NaCl + NaOCl + H 2O 2-July-2016 Chemsheets A Page 5

6 TASK 5 General AS Redx Questins 1) a) Write balanced half-equatins fr each f the fllwing xidatin r reductin prcesses. i) Br - t Br 2 iv) H 2SO 4 t S ii) N 2 t NO 3 v) NO 3 t NH 4 + iii) V 3+ t VO 2+ vi) BrO 3 t Br 2 (6) b) Write a redx equatin fr the reactin f H 2SO 4 with Br - ins, prducing S and Br 2, using the half-equatins frm (a). Als state which species is xidised, which is reduced, and the xidising and reducing agents. c) When cncentrated nitric acid is added t cpper metal, the cpper is xidised t xidatin state +2 and the nitric acid is reduced t nitrgen (IV) xide. Derive half-equatins and then write an equatin fr the reactin. (Ttal 12) 2) State whether the fllwing three reactins are redx reactins r nt. Fr thse that are redx reactins, clearly indicate any changes in xidatin state. a) Zn + 2 HCl ZnCl 2 + H 2 b) CuO + 2 HCl CuCl 2 + H 2O c) MnO HCl MnCl 2 + Cl H 2O d) Cl 2 + H 2O HCl + HOCl (7) (Ttal 7) 3) a) Explain, in terms f electrns, what happens t xidising and reducing agents in reactins. (1) b) Using these definitins, explain which species is xidised and which is reduced, and the xidising and reducing agent in reactin belw. (2) Mg + H 2SO 4 MgSO 4 + H 2 c) What is the xidatin state f each element in the fllwing species: Na 2SO 4 Na 2S 2O 3 Na 2SO 3 S 8 KClO 3 KClO KH Na 2O 2 Na 2O (9) (Ttal 12) 4) a) Fr each f the fllwing reductin r xidatin prcesses write a half equatin. i) cnversin f H + t H 2 ii) cnversin f SO t SO 3 iii) cnversin f H 2O 2 t O 2 iv) cnversin f IO 3 t I 2 v) cnversin f I t I 2 (5) b) Use yur half-equatins t write a redx equatin fr the reactin f IO - 3 with I - t frm I 2. (1) (Ttal 6) 5) Write redx equatins fr the fllwing reactins using the halfequatins prvided. MnO H e - Mn H 2O 2 Cl Cl e - 2 Cr H 2O Cr 2O H e a) Reactin f acidified MnO 4 - and Cl -. (1) b) Reactin f acidified MnO 4 - and Cr 3+. (1) (Ttal 2) 2-July-2016 Chemsheets A Page 6

7 SECTION 2 ELECTRODE POTENTIALS 1) The ptential f an electrde When a piece f metal (M) is dipped int a slutin f its metal ins (M n+ ), an equilibrium is set up. There is a tendency fr the metal t frm psitive ins and g int slutin. Hwever, there is als a tendency fr the metal ins in slutin t gain electrns and frm metal. M n+ (aq) + n e e.g. Zn 2+ (aq) + 2 e M(s) Zn(s) If this equilibrium lies t the left, then the metal acquires a negative charge due t a build up f electrns n the metal (the electrde has a negative ptential). If the equilibrium lies t the right, then a psitive charge builds up n the metal as electrns have been used up t frm metal frm the metal ins (the electrde has a psitive ptential). The psitin f the equilibrium (and s the charge n the metal) depends n the metal. Fr example, reactive metals tend t frm M n+ ins, s negative charge builds up n the metal (s reactive metals have negative ptentials). The mre unreactive metals tend t have psitive charge n the metal (s unreactive metals have psitive ptentials). A metal dipping int a slutin f its ins is called a half-cell r an electrde. There are ther types f half-cell where there is n slid metal invlved in the half-equatin. Fr these half-cells, a metal electrde is required and usually platinum is used as it is s unreactive (an inert electrde). Here are three general types f electrde. i) Metal electrdes These are the type met abve, which cnsist f a metal surrunded by a slutin f its ins, e.g. Zn(s) Zn 2+ (aq). ii) Gas electrdes iii) Redx electrdes This is fr a gas and a slutin f its ins. Here an inert metal (usually platinum) is the actual electrde t allw the flw f electrns, e.g. Pt(s) H 2(g) H + (aq) This is fr tw different ins f the same element (e.g. Fe 2+ and Fe 3+ ), where the tw types f ins are present in slutin with an inert metal electrde (usually Pt) t allw the flw f electrns. e.g. Pt(s) Fe 2+ (aq), Fe 3+ (aq) 2) Measuring the ptential f an electrde The actual ptential (E) f a half-cell cannt be measured directly. T measure it, it has t be cnnected t anther half-cell f knwn ptential, and the ptential difference between the tw half-cells measured. Cmbining tw half-cells tgether prduces an electrchemical cell. Befre the ptential f any half-cells culd be measured, a ptential had t be assigned t ne particular half-cell (then the ptential f all the ther electrdes culd be measured against it). The electrde chsen was the standard hydrgen electrde (SHE) and this electrde is assigned the ptential f 0 vlts. The SHE is knwn as the primary standard as it is the ptential t which all thers are cmpared. 2-July-2016 Chemsheets A Page 7

8 3) Setting up an electrchemical cell The tw half cells are jined tgether t give a cmplete circuit: the tw metals are jined with a wire (electrns flw thrugh the wire) the tw slutins are jined with a salt bridge (ins flw thrugh the salt bridge) a vltmeter is ften included in the circuit t allw the ptential difference (emf) t be measured A salt bridge is either a piece f filter paper saked with a slutin f unreactive ins r a tube cntaining unreactive ins in an agar gel Cmpunds such as KNO 3 are ften used in salt bridge as K + and NO 3 ins are quite unreactive. Salt bridge [strip f filter paper saked in saturated KNO 3(aq)] One f the tw half cells [a piece f zinc in a slutin f zinc nitrate(aq)] 4) Standard cnditins There are several factrs which affect the ptential f a half-cell, s they are measured under standard cnditins. cell cncentratin cell temperature cell pressure 1.0 ml dm -3 f the ins invlved in the half-equatin 298 K 100 kpa (nly affects half-cells with gases) The ptential shuld als be measured under zer-current cnditins [t measure the full ptential difference (emf), n current must be drawn frm the cell - this is achieved by using a high resistance vltmeter. A standard ptential is written as E. Standard cnditins are required because the psitin f the redx equilibrium will change with cnditins. Fr example, in the equilibrium: M n+ (aq) + n e M(s) an decrease in the cncentratin f M n+ wuld mve the equilibrium t the left, s making the ptential mre negative. 5) Standard hydrgen electrde (SHE) The ptential f all electrdes are measured by cmparing their ptential t that f the standard hydrgen electrde. H2 at 100 kpa salt bridge This is therefre called the primary standard as it is the standard t which all ther ptentials are cmpared. 2 H + (aq) + 2 e H 2(g) E = V The cell ntatin is: Pt(s) H 2(g) H + (aq) temperature = 298 K Pt 1.0 M H + (aq) 2-July-2016 Chemsheets A Page 8

9 TASK 6 The effect f changing cnditins n electrde ptential The standard ptential fr this half equatin is exactly 0.00 V: 2H + (aq) + 2e H 2(g) (H = -874 kj ml -1 ) a) What are the cnditins under which the ptential is 0.00 V? b) Why is the ptential exactly 0.00 V under these cnditins? c) In each case, state and explain if the value f the ptential fr this half equatin wuld becme mre r less than 0.00 V, if it changes at all, if the change stated was made: i) use hydrgen gas at 200 kpa ii) use a piece f platinum with a greater surface area iii) use H 2SO 4(aq) with cncentratin ml dm -3 iv) use HCl(aq) with cncentratin 2.00 ml dm -3 v) use a temperature f 50C 2-July-2016 Chemsheets A Page 9

10 6) Writing cnventinal representatins f cells (cell ntatin) Rather than drawing cmplicated diagrams f electrchemical cells, a shrthand frm is written. Tw examples are shwn belw. R O O R left hand half cell right hand half cell Cu(s) Cu 2+ (aq) Zn 2+ (aq) Zn(s) phase bundary phase bundary salt bridge V high resistance vltmeter left hand half cell right hand half cell H2 at 100 kpa salt bridge Pt(s) H 2 (g) H + (aq) Cu 2+ (aq) Cu(s) temperature = 298 K Pt Cu phase bundaries phase bundary 1.0 M H + (aq) 1.0 M Cu 2+ (aq) salt bridge where represents a phase bundary (i.e. between species in different states) represents a salt bridge Nte that the actual slid electrdes are written at the tw ends. If there is n slid metal in the half equatin, then a piece f platinum is used. Nte that the mst xidised species are near the salt bridge in the middle, i.e. the nes with the highest xidatin state (= O) (ROOR reduced, xidised, xidised, reduced). Nte the cnventin that the half cell with the mre psitive ptential is set up as the right hand electrde, and s cells are usually drawn that way (except when measuring ptentials against primary r secndary standards when the standard electrde is always the left hand electrde). 2-July-2016 Chemsheets A Page 10

11 TASK 7 Writing cnventinal representatins f cells Write the cnventinal representatin fr each f the fllwing cells. Zn 2+ (aq) + 2 e - Zn(s) Cu 2+ (aq) + 2 e - Cu(s) Al 3+ (aq) + 3 e - Al(s) Pb 2+ (aq) + 2 e - Pb(s)..... Cr 2O 7 2-2H + (aq) + 2e - H 2(g) 2- Cr 2 O 7 (aq) + 14H + + 6e - 2Cr 3+ (aq) + 7H 2O Cr 3+ hydrgen in Fe 2+ (aq) + 2e - Fe(s) H + Mn 2+ MnO 4 - (aq) + 8H + + 5e - Mn 2+ (aq) + 4H 2O.. Ni 2+ (aq) + 2e - Ni(s) 2H + (aq) + 2e - H 2(g) Ag + (aq) + e - Ag(s) Zn 2+ (aq) + 2e - Zn(s) July-2016 Chemsheets A Page 11

12 7) Measuring E versus the SHE Fr any cell, its emf (Eº cell) is the ptential f the right hand electrde minus the ptential f the left hand electrde: emf = Eº cell = Eº right - Eº left When finding the ptential f a half-cell under test, the standard electrde is always the left hand electrde, which in the case f the SHE gives: Eº cell = Eº test - Eº SHE Eº cell = Eº test Eº cell = Eº (Cu 2+ /Cu) = V Eº cell = Eº (Zn 2+ /Zn) = V Pt(s) H 2(g) H + (aq) Cu 2+ (aq) Cu(s) Pt(s) H 2(g) H + (aq) Zn 2+ (aq) Zn(s) 8) The use f secndary standards The SHE is difficult t use (partly as it invlves use f a gas, and ne that it flammable), s ften a different standard is used which is easier t use These ther standards are themselves calibrated against the SHE. This is knwn as using a secndary standard - i.e. a standard electrde that has been calibrated against the primary standard 9) The redx prcess Metal atms lse electrns at the ne electrde (xidatin), making the electrde negative, which travel thrugh the wire t the ther electrde, adding t ins t prduce metal atms (reductin). Remember: xidatin / ande / negative Remember als that usually the psitive electrde is the right hand electrde 2-July-2016 Chemsheets A Page 12

13 TASK 8 Electrde ptentials 1) Calculate the standard electrde ptential f the Na + /Na electrde given that when it was jined t the standard hydrgen electrde, the cell emf was 2.71 vlts. 2) Calculate the emf f a cell with the standard AgCl/Ag electrde (Eº = V) as the left hand electrde and the Fe 2+ /Fe (Eº = 0.44 V) electrde as the right hand ne. 3) a) Calculate the emf f a standard cell: Zn(s) Zn 2+ (aq) Pb 2+ (aq) Pb(s) given that: Zn 2+ (aq) + 2 e Zn(s) Eº = V Pb 2+ (aq) + 2 e Pb(s) Eº = V b) If the same cell was set up under standard cnditins except fr the cncentratin f the Pb 2+ (aq) ins which wuld be ml dm -3, hw wuld the emf cmpare t part (a). Explain yur answer. 4) Calculate Eº cell f the fllwing cells using the Eº values. a) Ni(s) Ni 2+ (aq) Sn 4+ (aq), Sn 2+ (aq) Pt(s) b) Pt(s) I (aq) I 2(s) Ag + (aq) Ag(s) c) Pt(s) Cl - (aq) Cl 2(g) Br 2(l), Br (aq) Pt(s) Ni 2+ (aq) + 2e Ni(s) Eº = V Sn 4+ (aq) + 2e Sn 2+ (aq) Eº = V I 2(s) + 2e 2 I aq) Eº = V Ag + (aq) + e Ag(s) Eº = V Br 2(l) + 2e 2 Br (aq) Eº = V Cl 2(g) + 2e 2 Cl (aq) Eº = V 5) Calculate the E f the Cu 2+ /Cu cuple given: Cu(s) Cu 2+ (aq) Cl 2(g) Cl (aq) Pt(s) Eº cell = V Cl 2(g) + 2e 2 Cl (aq) Eº = V 6) Calculate the standard electrde ptentials f the half-cells fr which the ptential is nt given. Write yur answer by writing the half equatin with its ptential. a) Mg(s) Mg 2+ (aq) Ti 3+ (aq), Ti 2+ (aq) Pt(s) Eº cell = V Mg 2+ (aq) + 2e Mg(s) Eº = V b) K(s) K + (aq) Mg 2+ (aq) Mg(s) Eº cell = V Mg 2+ (aq) + 2e Mg(s) Eº = V c) Pt(s) Hg(l) Hg 2Cl 2(aq), KCl(aq) Rb + (aq) Rb(s) Eº cell = V Hg 2Cl 2(aq) + 2e 2 Hg (l) + 2 Cl (aq) Eº = V 7) Fr each f the fllwing questins, i) Draw the cell ntatin fr the cell prduced when the tw half cells are jined via a salt bridge. ii) Calculate the cell emf. Remember the cell emf shuld be psitive, althugh it may nt be if the SHE is invlved. a) Cr 2+ (aq) + 2e Cr(s) Eº = V Zn 2+ (aq) + 2e Zn(s) Eº = V b) Cu 2+ (aq) + 2e Cu(s) Eº = V Fe 3+ (aq) + e Fe 2+ (aq) Eº = V c) MnO 4 (aq) + 8H + (aq) + 5e Mn 2+ (aq) + 4H 2O(l) Eº = V Cl 2(g) + 2e 2Cl (aq) Eº = V 2-July-2016 Chemsheets A Page 13

14 SECTION 3 USING THE ELECTROCHEMICAL SERIES 1) What is the electrchemical series The electrchemical series is a list f electrde ptentials in rder f decreasing (r increasing) ptential. Part f the electrchemical series is shwn belw. Very psitive ptentials mean that they are gd at attracting electrns (by taking them frm smething else which is xidised) it is the species n the left f these equatins that d this s they are the best xidising agents Very negative ptentials mean that they are gd at giving away electrns (by giving them t smething else which is reduced) it is the species n the right f these equatins that d this s they are the best reducing agents 2-July-2016 Chemsheets A Page 14

15 2) Using the electrchemical series The electrchemical series is a list f electrde ptentials in rder f decreasing (r increasing) ptential. At times, remember the glden rule that when tw half equatins are put tgether, the nce with the mre psitive ptential gets the electrns (makes sense as the psitive ne attracts the negative electrns. The mre psitive half equatin gains the electrns Example questin 1 What reactin will take place when the Cl 2/Cl half cell is jined t the Br 2/Br - half cell? Explain why this happens. Cl 2 + 2e 2 Cl E = V Br 2 + 2e 2 Br - E = V Cl 2 + 2e 2 Cl E = V this is mre psitive s gains electrns (and s ges in the frwards directin) Br 2 + 2e 2 Br - E = V this therefre ges backwards Cl 2 + 2e 2 Cl and 2 Br Br 2 + 2e s verall: Cl Br 2 Cl + Br 2 this happens as E (Cl 2/ Cl ) is mre psitive than E (Br 2/ Br ) and s Cl 2 gains electrns Example questin 2 The Cu 2+ /Cu and Zn 2+ /Zn half cells were jined. Zn e Zn E = V Cu e Cu E = V a) Write an equatin fr the chemical reactin that takes place. b) Explain why this reactin takes place. c) State which is the psitive electrde. d) State at which electrde xidatin takes place. e) Calculate the cell emf. 0 + electrde + electrde e V V Cu e Cu a) Reactins: Zn Zn e Cu e Cu Equatin: Zn + Cu 2+ Zn 2+ + Cu b) this happens as E (Cu 2+ /Cu) is mre psitive than E (Zn 2+ /Zn) and s Cu 2+ gains electrns c) psitive electrde = Cu 2+ /Cu 0.76 V Zn e Zn d) xidatin at negative electrde e) cell emf = V 2-July-2016 Chemsheets A Page 15

16 TASK 9 Using the electrchemical series 1) a) Write an equatin fr the reactin that wuld take place if the Zn 2+ (aq)/zn(s) and Ni 2+ (aq)/ni(s) half-cells were cnnected tgether. b) Explain why this reactin takes place. Zn e Zn E = V Ni e Ni E = V 2) What reactin wuld take place if a piece f silver and cpper was placed in a slutin cntaining a mixture f silver nitrate and cpper sulphate. Justify this using electrde ptential data. Ag + + e Ag E = V Cu e Cu E = V 3) Which f the species MnO 4 - (aq), Cr 2O 7 2- (aq) and Fe 3+ (aq) are able t liberate Cl 2 frm an acidic slutin f sdium chlride? Explain yur reasning using the electrde ptentials. MnO 4 + 5e + 8H + Mn H 2O E = V Cl 2 + 2e 2 Cl E = V Cr 2O e + 14H + 2Cr H 2O E = V Fe 3+ + e Fe 2+ E = V 4) Electrchemical cells are set by cmbining the tw half cells shwn. In each case: i) Give the cnventinal representatin f the cell. ii) Calculate the cell emf. iii) Identify the ande. iv) Write a balanced equatin fr the reactin that will take place in the cell. v) Explain why the reactin takes place. a) Br 2(l) + 2 e 2 Br - (aq) E = V 2 H + (aq) + 2 e H 2(g) E = V b) Cu 2+ (aq) + 2 e Cu(s) E = +0.34V Fe 3+ (aq) + e Fe 2+ (aq) E = V 5) Tw half cells are jined tgether. Each half cell cntains cpper nitrate slutin. The standard ptential fr Cu 2+ /Cu is vlts. Cu(s) Cu 2+ (aq) Cu 2+ (aq)cu(s) In the left hand half cell the cpper nitrate has a cncentratin f 1.00 ml dm -3 (with a ptential f V) and in the right hand half cell the cncentratin is ml dm -3. a) Will the ptential f the right hand half cell be the same as V, r mre r less than this value. Explain yur answer. b) Which electrde will be the ande? c) Which way will electrns flw in the cell? Explain yur answer. 6) Fr each f the fllwing questins, predict whether the reactin will take place r nt in aqueus slutin. Give clear reasns fr yur predictin. If the reactin des ccur, write an equatin fr the reactin. (Use the electrde ptentials n page 14). a) Will H + xidise Fe t Fe 2+? b) Will H + xidise Cu t Cu 2+? c) Will Cr 2O 2-7 /H + xidise Cl t Cl 2? d) Will MnO - 4 /H + xidise Cl t Cl 2? e) Will Mg reduce V 3+ t V 2+? 2-July-2016 Chemsheets A Page 16

17 SECTION 4 COMMERICAL CELLS Electrchemical cells can be used as a cmmercial surce f electrical energy. Imprtant types f cell include nnrechargeable, rechargeable and fuels cells. One great advantage f this is that it is a surce f prtable electricity. Nte that a battery is mre than ne cell jined tgether (e.g. a car battery is made frm six cells jined tgether) what the general public calls a battery shuld usually be called a cell. 1) Nn-rechargeable cells In these cells, the chemicals are used up ver time and the emf drps. Once ne r mre f the chemicals have been cmpletely used up, the cell is flat and the emf is 0 vlts. These cells cannt be recharged and have t be dispsed f after their single use. Zinc-carbn This is the standard, cheap nn-rechargeable cell but has a fairly shrt life 0.80 V Zn(NH 3) e Zn + 2NH V 2MnO 2 + 2H + + 2e Mn 2O 3 + H 2O cell emf = verall reactin during discharge = Alkaline This is a higher cst cell but has a lnger life 0.76 V Zn e Zn V MnO 2 + H 2O + e MnO(OH) + OH emf = verall reactin during discharge = 2) Rechargeable cells In rechargeable cells the reactins are reversible they are reversed by applying an external current and regenerate the chemicals. Lithium in Used in phnes, tablets, cameras, laptps, etc V Li + + CO 2 + e LiCO V Li + + e Li emf = verall reactin during discharge = verall reactin during re-charge = 2-July-2016 Chemsheets A Page 17

18 Lead-acid Make f six cells & used in cars V PbO 2 + 3H + + HSO 4 + 2e PbSO 4 + 2H 2O 0.36 V PbSO 4 + H e Pb + HSO 4 emf = verall reactin during discharge = ANODE (-ve) verall reactin during re-charge = Nickel-cadmium V NiO(OH) + 2H 2O + 2e Ni(OH) 2 + 2OH 0.88 V Cd(OH) 2 + 2e Cd + 2OH emf = verall reactin during discharge = verall reactin during re-charge = 3) Fuel cells Fuel cells are smewhat different t ther cells. They have a cntinuus supply f the chemicals int the cell and s neither run ut f chemicals nr need re-charging, (but they d need t have a cnstant supply f the required chemicals). The mst cmmn fuel cell is the hydrgenxygen fuel cell. tubes bringing hydrgen and xygen t the fuel cell 2-July-2016 Chemsheets A Page 18

19 The diagram utlines at a simple level hw a hydrgen-xygen fuel cell wrks. H 2 IN ANODE (-ve) H 2 2H + + 2e e - electrns mve between electrdes thrugh a wire e - ELECTROLYTE H + ins mve between electrdes thrugh the electrlyte e - e - CATHODE (+ve) O 2 + 4H + + 4e 2H 2O e - O 2 IN H 2O OUT The hydrgen-xygen fuel cell can be run in alkaline r acidic cnditins, but the verall equatin and the verall emf is the same. hydrgen fuel cell (alkaline) hydrgen fuel cell (acidic) Negative electrde (ande) Psitive electrde (cathde) H 2 + 2OH 2H 2O + 2e E = 0.83 V H 2 2H + + 2e E = V O 2 + 2H 2O + 4e 4OH E = V O 2 + 4H + + 4e 2H 2O E = V Overall equatin 2H 2 + O 2 2H 2O 2H 2 + O 2 2H 2O Cell emf V V Fuel cells are very efficient and nly give ff water as a waste prduct. sme vehicles are pwered by hydrgen fuel cells 2-July-2016 Chemsheets A Page 19

20 4) Benefits and risks f using cells Benefits Risks Using cells prtable surce f electrical energy waste issues Using nn-rechargeable cells cheap waste issues Using re-chargeable cells Using hydrgen fuel cells less waste cheaper in the lng run lwer envirnmental impact nly waste prduct is water d nt need re-charging very efficient sme waste issues (at end f useful life) need cnstant supply f fuels hydrgen is flammable & explsive hydrgen usually made using fssil fuels high cst f fuel cells TASK 10 Cmmercial cells 1) Sme nn-rechargeable cells are zinc/silver xide cells. Ag 2O(s) + 2H + (aq) + 2e 2Ag(s) + H 2O(l) E = V Zn 2+ (aq) + 2e Zn(s) E = V a) Calculate the verall emf fr this cell. b) Write an equatin fr the reactin taking place when the cell is being used. c) Write the cnventinal representatin f this cell. d) Why wuld the cell emf f this cell fall ver time and eventually reach zer? 2) The nickel-metal hydride cell is a rechargeable cell. The half equatins are shwn belw where M represents a metal. The actual electrde ptential f the negative electrde and the verall emf varies depending n the metal. negative electrde M + H 2O + e MH + OH psitive electrde NiO(OH) + H 2O + e Ni(OH) 2 + OH a) Write an equatin fr the reactin taking place when the cell is being used. b) Write an equatin fr the reactin taking place when the cell is recharged. c) What is the main advantage f a rechargeable cell ver a nn-rechargeable cell. 3) Methanl fuel cells are an alternative t hydrgen fuel cells. O 2 + 4H + + 4e 2H 2O E = V CO 2 + 6H + + 6e CH 3OH + H 2O E = +0.02V a) Write an equatin fr the reactin taking place when the cell is being used. b) Calculate the verall emf f the cell. c) What is the main advantage f a fuel cell ver rechargeable and nn-rechargeable cells. 2-July-2016 Chemsheets A Page 20