Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
_____________________________
Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
The Ti 2+ ions is represented by electron configuration (Ar)3d2. Titanium is in atomic number 22 and its electronic configuration is (Ar)3d2 4s2. Titanium loss two electron that is 4s2 electrons hence the electronic configuration ( Ar)3d2. 4s2 is the valence electron hence it the one which is lost to form Ti2+
To solve this problem we just need to use the rule of three:
150g..................395.1J
450g................xJ
x = 450*395.1/150 = 1185,3J
450.0 g of the substance completely reacted with oxygen will produce 1.1853 kJ(<span>kiloJoule</span>)
