Answer: Ecell = -0.110volt
Explanation:
Zn--->Zn^+2 + 2e^-.........(1) oxidation
Cu^2+ 2e^- --->Cu........(2)reduction
Zn + Cu^2+ ----> Cu + Zn^+2 (overall
For an electrochemical cell, the reduction potential set up is given by
E(cell) = E(cathode) - E(anode)
E(cell) = E(oxidation) - E(reduction)
E(cathode) = E(oxidation)
E(anode) = E(reduction)
Given that
E(oxidation) = -0.763v
E(reduction) = +0.337v
E(cell) = -0.763 - (+0.337)
E(cell) = -0.763- 0.337
E(cell) = -0.110volt
Answer:
2 M Al³⁺(aq) and 6 M NO₃⁻(aq)
Explanation:
Al(NO₃)₃ is a strong electrolyte that ionizes according to the following equation.
Al(NO₃)₃(aq) → Al³⁺(aq) + 3 NO₃⁻(aq)
The solution 2.0 M contains 2.0 moles of Al(NO₃)₃ per liter of solution.
The molar ratio of Al(NO₃)₃ to Al³⁺ is 1:1. The concentration of Al³⁺ is:
2.0 mol Al(NO₃)₃/L × (1 mol Al³⁺/1 mol Al(NO₃)₃) = 2.0 mol Al³⁺/L = 2.0 M Al³⁺
The molar ratio of Al(NO₃)₃ to NO₃⁻ is 1:3. The concentration of NO₃⁻ is:
2.0 mol Al(NO₃)₃/L × (3 mol NO₃⁻/1 mol Al(NO₃)₃) = 6.0 mol NO₃⁻/L = 6.0 M NO₃⁻
Like PERIODic table.). All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons. All of the elements in the second row (the second period) have two orbitals for their electrons.