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
Answer:
1)Na2O
let the valency of Na is x
2(x)+(2)=0
2x+2=0
2x=-2
x=-1
2)ZnO
let the valency of Zn is x
x+2=0
x=-2
3)Al2O3
let the valency of Al is x
2(x)+3(2)=0
2x+6=0
2x=-6
x=-3
4)MgO
let the valency of Mg is x
x+2=0
x=-2
D) basic ions contain some H+ ions
