Read the passage given below to answer questions:

A potential difference developed between electrode and electrolyte is called electrode potential. When the concentrations of all the species involved in a half cell is unity, then electrode potential is known as standard electrode potential. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential with respect to solution. The other half-cell in which reduction takes place is called cathode and it has a positive potential with respect to solution. Thus, there exists a potential difference between two electrodes, cathode and anode. This difference is called cell potential and is measured in volts. It is called the cell electromotive force when no current is drawn through the cell. A galvanic cell is represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by salt bridge. Under this convention, emf of cell is positive and is given as

E_cell = E_right - E_left

The standard electrode potential is very important. The value at standard electrode potential of an electrode is greater than zero, then its reduced form is more stable compared to hydrogen gas. The value at some standard electrode potentials at 298 K are given below. (Ions are present as aqueous species and H₂O as liquid)

The EMF of the cell Ag(s) | Ag⁺ (1m) || Pb²⁺ (1m) | Pb(s), is

-0.93 V

Ag(s)|Ag⁺(1m) || Pb²⁺(1m)|Pb(s) 

At anode: 2Ag → 2Ag⁺ + 2e^-   

At cathode: Pb²⁺ + 2e^- → Pb

2Ag + Pb²⁺ → 2Ag⁺ + Pb

E^o_cell = E^o_Pb²⁺/Pb - E^o_Ag⁺/Ag

E^o_cell = -0.13 - 0.80 = -0.93 V

E_cell = E^o_cell - \(\frac{0.059}{n}\)log\(\frac{[Ag^+]^2}{[Pb^{2+}]}\)

E_cell = -0.93 - \(\frac{0.059}{2}\)log\(\frac{[1]^2}{[1]}\)     [(log1 = 0)]

E_cell = -0.93 V