Answer:
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.
