Consider a galvanic cell composed of the SHE and a half-cell using the following reaction: Ag + (aq) + e - → Ag (s)

(a) Calculate the standard cell potential. E o cell = V

(b) What is the spontaneous cell reaction under standard-state conditions?

(c) Calculate the cell potential when [H+] in the hydrogen electrode is changed to the following concentrations, while all other reagents are held at standard-state conditions:

(i) 4.2 * 10-2 M E = V

(ii) 9.6 * 10-5 M E = V

(d) Based on this cell arrangement, suggest a design for a pH meter.

a. 0.80V b. 2Ag⁺ (aq) + H2 (g) ⇄ 2Ag (s) + 2H⁺ (aq) c. i) 0.88 ii) 1.03 d. Cell is a ph meter with the potential being a function of hydrogen ion concentration

Explanation:

a. The two half cell reactions are

1. 2H⁺ (aq) + 2e⁻ → H₂ (g) Eanode = 0.00V

2. Ag⁺ (aq) + e⁻ → Ag (s) Ecathode = 0.80V

The balanced cell reaction is

2Ag (aq) ⁺ + H₂ (g) ⇄ 2Ag (s) + 2H⁺ (aq)

therefore Ecell = Ecathode - Eanode = 0.80 - 0.00 = + 0.80V

b. Since the Ecell is positive, the spontaneous cell reaction under standar state conditions is

2Ag (aq) ⁺ + H₂ (g) ⇄ 2Ag (s) + 2H⁺ (aq)

c. Use Nernst Equation

E = Ecell - (0.0592/n) log ([H⁺]/[Ag⁺]²[P H₂]), where n is the number of moles of Ag and P H₂ = 1.0 atm

i) E = 0.80 - (0.0592/2) log (4.2x10^-2) / (1.0) ² (1.0) = 0.88V

ii) E = 0.80 - (0.0592/2) log (9.6x10^-5) / (1.0) ² (1.0) = 1.03V

d. From the above calculation we can conclude that the cell acts as a pH meter as a change in hydrogen ion concentration results in a change in the potential of the cell. A change of ph of 2.64 changes the E of cell by 0.15 V.