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Form
                                                                                                           5
                                                                   Chapter 1 Redox Equilibrium  Chemistry
                                                                                                  +
 CHAP.                                                             (b)  This  means hydrogen ions, H  undergo  CHAP.
 1                                                                     reduction  to  form  hydrogen  gas,  H .   1
                                                                                                          2
                   The standard reference half-cell is the standard    Therefore hydrogen electrode acts as the
                   hydrogen electrode, SHE.                            cathode.
                                                                         2H O (aq) + 2e  → H (g) + 2H O(l)
                                                                                     –
                                                                             +
                                                                           3
                                                                                          2
                                                                                                  2
                   2.  An  unknown standard electrode potential,                 or simplified as
                     E 0 unknown  can be obtained by constructing a           2H (aq) + 2e  → H (g)
                                                                                        –
                                                                                +
                                                                                              2
                     voltaic cell consisting of a  reference half-cell   (c)  On  the other hand, zinc atoms, Zn are
                     and another unknown half-cell, to measure the     oxidised to zinc ions, Zn . Hence, zinc
                                                                                              2+
                     electromotive force, e.m.f or commonly known      electrode acts as the anode.
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                     as standard cell potential, E 0 cell .            Oxidation half reaction:
                     (a)  If the hydrogen ion, H  is  reduced, the              Zn(s) → Zn (aq) + 2e –
                                             +
                                                                                          2+
                        reference half-cell acts as the cathode while   (d)  E 0  cell   = E 0 cathode  – E 0 anode
                        oxidation occurs at the unknown half-cell.      0.76 V = 0.00 V – E 0

                                                                                        zinc
                        E 0   = E 0   – E 0                             E 0     = 0.00 V – 0.76 V
                                       anode
                               cathode
                          cell  = 0.00 V – E 0 unknown                      zinc  = – 0.76 V
                            = – E 0                                     Zinc standard electrode potential:
                                 unknown
                        Standard electrode potential of the half-cell:  Zn  (aq) + 2e   Zn(s)   E 0 zinc  = – 0.76 V
                                                                                    –
                                                                           2+
                        E 0 unknown  = –E 0 cell
                     (b)  If  the  hydrogen  gas,  H   is  oxidised,  the
                                             2
                        reference half-cell acts as the  anode while
                        reduction occurs at the unknown half-cell.  Conventionally, standard electrode potential is
                                                                  written in the form of standard reduction potential.
                        E 0 cell  = E 0 cathode  – E 0 anode
                            = E 0 unknown  – 0.00 V
                            = E 0 unknown                        Oxidising Agents and Reducing Agents Based on
                        Standard electrode potential of the half-cell:  The Value of Standard Electrode Potentials
                                                                                                          0
                        E 0     = E 0                             1.  The  value of standard electrode potential, E
                          unknown  cell
                      Example:                                      gives a direct measure of the ease of a chemical
                      Figure 1.13 shows a voltaic cell consisting of a   species to be oxidised or reduced.
                     standard zinc half-cell and a standard hydrogen
                     half-cell which is prepared to determine the
                     standard electrode potential of a zinc.          Chemical species can be atom, molecule,
                                   Voltmeter                          monoatomic ion, polyatomic ion, or radical.
                              e –    0.76 V  e –
                                                                  2.  We can compare the strength in oxidation and
                                 Anode (–)   Cathode (+)            reduction of elements or ions based on the value
                               Zn                H (g)              of E  for the half reactions involved.
                                                                       0
                                                  2
                                                                   (a)  The more positive the E  value is, the easier
                                                                                           0
                                                     H                 for the chemical species on the left side of the
                                                      2
                   e –    Zn 2+         Pt       2H O                  half equation to undergo reduction.
                                                   2
                                                                                           0
                    e –                           2H O +  e –      (b)  The more negative the E  value is, the easier
                       Zn      1 M Zn 2+  1 M H O +  3
                                                                       for the chemical species on the right side of
                                            3
                                   Salt bridge                         the half equation to undergo oxidation.
                                  Figure 1.13                      Example:
                     (a)  There are a few observations that can be seen        Figure 1.14  shows the standard electrode
                        in the voltaic cell:                        potentials of three ions.
                         (i)  Gas bubbles are produced around the
                             platinum, Pt electrode.                   Zn  (aq) + 2e  ⇌ Zn (s)  E  = –0.76 V
                                                                                           0
                                                                                 –
                                                                         2+
                                                                                           0
                                                                                 –
                                                                          +
                         (ii)  Zinc, Zn electrode becomes thinner.  Strength as   oxidising   agent  2H  (aq) + 2e  ⇌ H  (g)  E  = 0.00 V  Strength as   reducing   agent
                                                                                     2
                                                                         2+
                                                                                 –
                                                                                           0
                         (iii)  Voltmeter reading shows 0.76 V.        Cu  (aq) + 2e  ⇌ Cu (s)  E  = 0.34 V
                                                                                 Figure 1.14
                 1.2.1 1.2.2                                                                          313
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