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.

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.