(i) The formation of rust requires both oxygen and water. Since it is an electrochemical redox process, it requires an anode and cathode in different places on the iron.
(ii) The iron surface and a droplet of water on the surface form a tiny galvanic cell.
(iii) The region enclosed by water is exposed to low amount of oxygen and it acts as the anode.
(iv) The remaining area has high amount of oxygen and it act a cathode. So an electro chemical cell is formed. Corrosion occurs at the anode i.e., in the region enclosed by the water.

Anode (oxidation)Iron dissolve in the andoe region:
\({ 2Fe }_{ (s) }\rightarrow { 2Fe }_{ (aq) }^{ 2+ }+{ 4e }^{ - }\) E^o = 0.44V
The electrons move through the iron metal from the anode to the cathode area where the oxygen dissolved in water, is reduced to water.
Cathode (reduction):
The reaction of atmospheric carbon dioxide with water gives carbonic acid which furnishes the H⁺ ions for reduction.
\({ O }_{ 2(g) }+{ 4H }_{ (aq) }^{ + }+{ 4e }^{ - }\rightarrow { 2H }_{ 2 }O(l)\)  E^o = 1.23 V
The electrical circuit is completed by the migration of ions through water droplet.
The overall redox reactions is,
\({ 2Fe }_{ (s) }+{ O }_{ 2(g) }+{ 4H }_{ (aq) }^{ + }\rightarrow { 2Fe }_{ (aq) }^{ 2+ }+{ 2H }_{ 2 }O(l)\)  E^o = 0.444 + 1.23 = 1.67V
The positive emf value indicates that the reaction is spontaneous
Fe²⁺ ions are further oxidised to Fe³⁺ which on further reaction with oxygen to form rust.
\({ 4Fe }_{ (Aq) }^{ 2+ }+{ O }_{ 2(g) }+{ 4H }_{ (aq) }^{ + }\rightarrow { 4Fe }_{ (aq) }^{ 2+ }+{ 2H }_{ 2 }O(l)\)
\({ 3Fe }_{ (Aq) }^{ 3+ }+{ 4{ H }_{ 2 }O }(l)\rightarrow { Fe }_{ 2 }{ O }_{ 3 }.{ H }_{ 2 }{ O }_{ (s) }+{ 6H }_{ (aq) }^{ + }\)

Standard Hydrogen Electrode (SHE) is used as the reference electrode. It has been assigned an arbitrarily emf of zero volt. It consists of a platinum electrode in contact with 1M HCI solution and 1atm hydrogen gas. The hydrogen gas is bubbled through the solution at 25°C. SHE can act as a cathode as well as an anode. The Half cell reactions are given below.
If SHE is used as a cathode, the reduction reactions is
2H⁺ (aq,1M) + 2e^- ⟶ H₂ (g, 1 atm)  E^o= 0 volt
If SHE is used as an anode, the oxidation reaction is
H² (g.1 atm) ⟶ 2H⁺ (aq, 1M) + 2e^- E^o= volt
Illustration: Let us calculate the reduction potential of zinc electrode dipped in zinc sulphate solution using SHE.
Step 1: The following galvanic cell is constructed using SHE
Zn_(s) | Zn²⁺ (aq,1M) || H⁺ (aq, 1M) | H₂ (g, 1 atm)|pt_(s)
Step 2: The emf of the above galvanic cell is measured using a volt meter. In this case, the measured emf of the above galvanic cell is 0.76V.
Calculation
We know that,
E^o_cell = (E^o_ox)_Zn|Zn²⁺ + (E^o_red)_SHE
E^o_cell = 0.79 and (E^o_red)_SHE = 0V.
Substitute these values in the above equation
⇒ 0.76V = (E^o_ox) _Zn|Zn2+ + 0V
⇒ (E^o_ox)_Zn|Zn²⁺ = 0.76V
This oxidation potential corresponds to the below mentioned half cell reaction which takes place at the cathode.
Zn ⇾ Zn²⁺ + 2e^- (Oxidation)
The emf for the reverse reaction will give the reduction potential
Zn²⁺ + 2e^- ⇾ Zn; E^o= - 0.76V
∴ (E^o_ox)_Zn²⁺|Zn = - 0.76V

(i) The conductivity of an electrolytic solution is determined by using a wheatstone bridge arrangement in which one resistance is replaced by a conductivity cell filled with the electrolytic solution of unknown conductivity.
(ii) In the measurement of specific resistance of a metallic wire, a DC power supply is used. Here, if we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell. So, AC current is used for this measurement to prevent electrolysis

(iii) A wheatstone bridge is constituted using known resistances P, Q, a variable resistance S and conductivity cell (Let the resistance of the electrolytic solution taken in it be R) as shown in the figure. An AC source (550 Hz to 5 KHz) is connected between the junctions A and C. Connect a suitable detector (Such as the telephone ear piece detector) between the junctions 'B' and 'D'.
(iv) The variable resistance (S) is adjusted until the bridge is balanced and in this conditions there is no current flow through the detector.
 Under balanced condition,
\(\frac { P }{ Q } =\frac { R }{ S } \)
\(\therefore R=\frac { P }{ Q } \times S\)
(v) The resistance of the electrolytic solution (R) is calculated from the known resistance values P, Q and the measured 'S' value under balanced condition using the above expression

E^o_cell = 1.56V [Zn²⁺]
= 0.1 M [Ag⁺] = 10 M
Formula:
\([{ E }_{ cell }={ E }_{ cell }^{ o }-\frac { RT }{ nF } In\frac { [{ Zn }^{ 2+ }] }{ [{ Ag] }^{ 2 } } ]\)
Solution:
= 1.56 - 0.02955 log 0.001;
= 1.56 - (- 0.08865)
= 1.56 + 0.08865 = 1.6486 V
E_cell = 1.6486 V.

Given:
Zn/ZnO₂, OH^-, HgO/Hg
E^o_R = 0.098 V; E_L= -1.216 V
Formula: ∴ E^o_cell = E^o_R-E^oL
Solution: E^o_cell = 0.098-(-1.216)
= 0.098 + 1.216 = +1.314
E^o = + 1.314 V.