Page 7 - 1202 Question Bank Chemistry Form 5 KSSM
P. 7

(d)  Three  observations  may  be  observed  during  a     5.  The standard cell potential can be obtained based on
                      displacement reaction of a metal.             the following equation.
                      •  The more reactive metal dissolves
                      •  The less reactive metal is deposited                 E 0 cell  = E  0 cathode  − E 0 anode
                      •  The colour of the salt solution may change
                                                                                              0
                                                                  6.  The electrode potential value, E  is used to predict:
                9.  Displacement of halogen from its halide solution  (a)  Substance that undergoes oxidation or reduction.
                  (a)  Halogens  are  oxidising  agents.    Conversely,   (b)  Substance that acts as an oxidising or reducing
                      halide ions are reducing agents.                  agent.
                  (b)  A halogen molecule, X , gains electrons to form
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                                        2                           (c)  Strength of oxidising or reducing agents.
                                   −
                                               −
                      the halide ion, X  : X (aq) + 2e  → 2X (aq)
                                                    −
                                       2                                Example:
                  (c)  Reactivity  or  oxidising  power  of  the  halogens
                      decreases going down Group 17.                              F    + 2e    2F −  E  = +2.87 V
                                                                                                    0
                                                                                        −
                  (d)  The  halide  ion  gains  electrons  to  form  the           2
                                          −
                                                  −
                      halogen molecule, X : 2X (aq) + 2e  → X (aq)
                                                                                                    0
                                                                                         −
                                      2                2                Oxidising   CI   + 2e    2CI −  E  = +1.36 V
                  (e)  Reducing  power  of  the  halide  ions  increases   agent    2
                      moving down Group 17.
                                                                                  Br   + 2e    2Br −  E  = +1.07 V
                                                                                                    0
                                                                                         −
                                                                                    2
               1.2   Standard Electrode Potential
                                                                        •  E° value increases, strength as oxidising agent
                                               0
                1.  Standard electrode potential, E  is defined as       increases.
                                                                        •  Increasing order of oxidising agent strength:
                  the  difference  of  electrode  potential  (voltage)  of  an   Br , Cl , F
                                                                           2  2  2
                  electrode  system  consisting  of  an  electrode  half-cell
                  pairing up with the standard hydrogen electrode (SHE)
                  half-cell.                                     1.3   Voltaic Cell
                                        Voltmeter
                                      V                           1.  A simple voltaic cell consists of two different metals
                                                                    immersed in an electrolyte solution and connected
                           H (g)
                            2     Salt bridge                       with connecting wires.
                 298 K and 1 atm              Electrode X
               Platinum electrode                                 2.  A Daniell cell is an example of a voltaic cell where
               Acid solution                                        zinc metal and copper metal are used as electrodes
               (concentration of              Solution of metal X   dipped into their respective ionic salt solutions.
               H  is 1.0 mol dm )  SHE        ion 1.0 mol dm –3
                 +
                           –3
                                                                                    Voltmeter
                                                                                 e –       e –
                2.  The  standard  condition  to  measure  the  standard
                                   0
                  electrode potential, E  of the cell:                             Salt bridge
                  (a)  Concentration of ion in aqueous solution is 1.0   Anode  (–)  Zn   Cu  (+)  Cathode
                            −3
                      mol dm .                                          (Oxidation)   KCl      (Reduction)
                  (b)  Gas pressure of 101 kPa or 1 atm.
                  (c)  Temperature at 298 K or 25 °C.
                  (d)  Platinum is used as an inert electrode.
                                                                                ZnSO     CuSO
                                                                                    4        4
                       0
                3.  The  E   value  of  standard  hydrogen  electrode,  SHE
                                                                           2+
                                                                                            0
                                                                                −
                  = 0.00 V.                                              Zn  + 2e    Zn     E  = −0.76 V
                                                                         Cu  + 2e    Cu    E  = +0.34 V
                                                                           2+
                                                                                            0
                                                                                −
                                  1
                                             0
                              –
                      H (aq) + e  ⇌    H (g)       E  = 0.00 V
                       +
                                                                              2+
                                  2  2                              Zn(s) → Zn (aq) + 2e (Oxidation)
                                                                                      −
                                                                       2+
                                                                               −
                                                                    Cu (aq) + 2e  → Cu(s) (Reduction)
                                                                    Overall ionic equation:
                4.  The above cell can be represented in the form of a cell   2+      2+
                  notation.                                         Zn(s) + Cu (aq) → Zn (aq) + Cu(s)
                                                                    Cell notation of Daniell cell:
                                        n+
                    Pt(s) | H (g) | H (aq)  ||  X (aq) | X(s)
                               +
                                                                        2+
                                                                                          2+
                                                                                                       −3
                                                                                     −3
                          2                                     Zn(s) | Zn (aq, 1.0 mol dm ) || Cu (aq, 1.0 mol dm ) | Cu(s)
                     ⎧  ⎪  ⎪  ⎨  ⎪  ⎪  ⎩  ⎧  ⎪  ⎨  ⎪  ⎩
                   •  Represents the   •  Represents electrode      Cell voltage, E 0    = E 0   − E 0
                     SHE                X                                        cell  0 cathode  0  anode
                                                                                   = E   − E
                                                                                      Cu    Zn
                                                                                   = 0.34 − (−0.76)
                                                                                   = +1.10 V
              2
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