Question

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.

Answer 1

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

Step-by-step 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.