Galvanic Cell (Part 2)

1. Standard reduction/electrode potential (E^o)
    (a) Electrons in different types of metals have different potential energy. Electrode potential (E) is a measure of the half-cell to attract electrons. Its unit is volt (V).
    (b) Standard electrode potential (E^o) is a measure of the half-cell to attract electrons at standard conditions.
    Example:
    Cu²⁺ (aq) + 2e^- → Cu (s)        E^o = +0.34 V
    Zn²⁺ (aq) + 2e^- → Zn (s)         E^o = -0.76 V
    (c) Standard conditions refer to conditions where the temperature is 25^oC, the pressure is 1 atm (for gases), and the concentration of ions in the electrolyte is 1.0 mol dm^-3.

2. Standard cell potential (E^ocell)
    (a) The potential difference between the two metals acts a driving force or ‘electrical pressure’ that pushes electrons through the wire. The potential difference is termed cell potential.
    (b) Standard cell potential (E^ocell) is the difference between the standard reduction potential of the cathode and the standard reduction potential at the anode.
        E^ocell = E^ocathode - E^oanode
    @ E^ocell = E^ored + E^oox

    Example:
    Calculate the standard cell potential for the following reactions:
        Cu (s) + 2 Ag⁺ (aq) → Cu²⁺ (aq) + 2 Ag (s)

    Method 1
    Cu²⁺ (aq) + 2e^- → Cu (s)        E^o = +0.34 V
    Ag⁺ (aq) + e^- → Ag (s)           E^o = +0.80 V
   
    From the overall cell reaction equation, it shows that copper half-cell is the anode (since it undergoes oxidation: an increase in oxidation number) and silver half-cell is the cathode (since it undergoes reduction: a decrease in oxidation number).
    E^ocell = E^ocathode - E^oanode
            = +0.80 - (+0.34)
            = +0.46 V

    Method 2
    Cu (s) → Cu²⁺ (aq) + 2e^-            E^o = -0.34 V
    2 Ag⁺ (aq) + 2e^- → 2 Ag (s)        E^o = +0.80 V
    Cu (s) + 2 Ag⁺ (aq) → Cu²⁺ (aq) + 2 Ag (s)
   
    E^ocell = E^ored + E^oox
            = +0.80 + (-0.34)
            = +0.46 V

    (c) The electrode potential of zinc half-cell written as a reduction reaction is -0.76 V. When the zinc half-cell equation is written as an oxidation reaction, the sign of its E^o value is reversed. When a half-equation is written in reverse, the sign is reversed too, but the value remains the same.
        Zn²⁺ (aq) + 2e^- → Zn (s)        E^o = -0.76 V
        Zn (s) → Zn²⁺ (aq) + 2e^-        E^o = +0.76 V

    (d) The reduction half-cell reactions may be written with different coefficients for the species involved, but the values and signs of reduction potentials do not change.
        ½ Cl2 (g) + e^- → Cl (aq)         E^o = +1.36 V
        Cl2 (g) + 2e^- → 2 Cl^- (aq)       E^o = +1.36 V

3. Relative strength of oxidising agents or reducing agents
    (a) Electrode potentials may be positive or negative. A half-cell with a positive electrode potential is able to attract electrons towards it. The more positive a half-cell’s electrode potential, the stronger is its attraction for electrons.
    (b) Species on the left-hand side of a reduction half-equation with a positive standard reduction potentials is an oxidising agent. The oxidising power is greater when the value of the reduction potential is more positive.
    (c) Species on the right-hand side of a reduction half-equation with a negative standard reduction potentials is a reducing agent. The reducing power is greater when the value of the reduction potential is more negative.
    Example:
    Ag⁺ (aq) + e^- → Ag (s)           E^o = +0.80 V
    Cu²⁺ (aq) + 2e^- → Cu (s)        E^o = +0.34 V
    Ni²⁺ (aq) + 2e^- → Ni (s)          E^o = -0.25 V

    Increasing strength of oxidising agent: Ni²⁺ < Cu²⁺ < Ag⁺
    Increasing strength of reducing agent: Ag < Cu < Ni

4. Standard hydrogen electrode (SHE)
    Standard hydrogen electrode (SHE) is made up of a platinum electrode coated with finely divided platinum (to act as catalyst for the half-cell reaction), immersed in an aqueous solution of H⁺ ions (normally hydrochloric acid) of concentration 1.0 mol dm^-3 and bubbled with hydrogen gas at 1 atm pressure. The temperature is kept constant at 25^oC.
    

    (a) The standard reduction potential of SHE is assigned exactly 0 V.
    (b) The direction of the half-reaction of SHE depends on the other half-cell connected to it. SHE serves as a reference to determine the standard electrode (reduction) potential of other half-cell.

5. Determination of standard reduction electrode of zinc (Zn) half-cell
    

    (a) The voltmeter reading gives +0.76 V with a deflection of the needle indicating electrons are flowing from zinc electrode to SHE.
    (b) The half-cell equations and overall cell reaction equation
        Anode:     Zn (s) → Zn²⁺ (aq) + 2e^-
        Cathode:  2 H⁺ (aq) + 2e^- → H2 (g)
        Overall reaction:  Zn (s) + 2 H⁺ (aq) → Zn²⁺ (aq) + H2 (g)

    (c) Cell notation
        Zn (s) | Zn²⁺ (aq) || H⁺ (aq) | H2 (g) | Pt (s)

    (d) Standard reduction potential of zinc half-cell
        E^ocell = E^ocathode - E^oanode
        +0.76 = 0.00 - E^oZn²⁺/Zn
        E^oZn²⁺/Zn = -0.76 V

6. Determination of standard reduction electrode of copper(Cu) half-cell
   

    (a) The voltmeter reading gives +0.34 V with a deflection of the needle indicating electrons are flowing from SHE to copper electrode.
    (b) The half-cell equations and overall cell reaction equation
        Anode:     H2 (g) + 2e^- → 2 H⁺ (aq)
        Cathode:  Cu (s) → Cu²⁺ (aq) + 2e^-
        Overall reaction: H2 (g) + Cu (s) → 2 H⁺ (aq) + Cu²⁺ (aq)
   
    (c) Cell notation
        Pt (s) | H2 (g) | H⁺ (aq) || Cu (s) | Cu²⁺ (aq)

    (d) Standard reduction potential of copper half-cell
        E^ocell = E^ocathode - E^oanode
        +0.34 = E^oCu²⁺/Cu - 0.00
        E^oCu²⁺/Cu = +0.34 V

7. Spontaneity of a redox reaction
    A redox reaction is spontaneous when its cell potential (Ecell) is positive. Conversely, a redox reaction is non-spontaneous when its cell potential is negative.

    Example:
    Predict the spontaneity of the following reaction under standard conditions.
    Zn (s) + Sn⁴⁺ (aq) → Zn²⁺ (aq) + Sn²⁺ (aq)
    [E^oZn²⁺/Zn = -0.76V; E^oSn⁴⁺/Sn²⁺ = +0.13 V]

    Method 1
    From the overall cell reaction equation, it shows that zinc half-cell is the anode (since it undergoes oxidation: an increase in oxidation number) and tin half-cell is the cathode (since it undergoes reduction: a decrease in oxidation number).
    E^ocell = E^ocathode - E^oanode
            = +0.13 - (-0.76)
            = +0.89 V
    Thus, the redox reaction is spontaneous since the E^ocell is positive.

    Method 2
    Zn (s) → Zn²⁺ (aq) + 2e^-          E^o = +0.76 V
    Sn⁴⁺ (aq) + 2e^- → Sn²⁺ (aq)    E^o = +0.13 V
    Zn (s) + Sn⁴⁺ (aq) → Zn²⁺ (aq) + Sn²⁺ (aq)

    E^ocell = E^ored + E^oox
            = 0.13 + 0.76
            = +0.89 V
    Thus, the redox reaction is spontaneous since the E^ocell is positive.