The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
Answer:
85.5%
Explanation:
To get the experimental value, you need to convert 15.0 grams of Na2SO4 to grams of Na3PO4. You do this with stoichiometry.
Convert grams of Na2SO4 to moles with molar mass. Then convert to moles of Na3PO4 with the mole-to-mole ratio according to the balanced chemical equation. Then convert moles of Na3PO4 to grams with the molar mass.
15.0 g Na2SO4 x (1 mol/142.04 g) x (2 Na3PO4/3Na2SO4) x (163.94 g/1 mol) = 11.7 g Na3PO4
Percent Yield = (actual value/experimental value) x 100
Actual Value = 10.0 g
Experimental Value = 11.7 g
10.0g/11.7 g = 85.5%
3Mg(s) + 2FeCl3(aq) → 3 MgCl2(aq) + 2Fe(s)
Explanation:
Redox reaction is a reduction-oxidation reaction in which oxidation states of atoms involved in the reaction are changed. The elements that receive electrons are considered to be reduced while the element that donates the electrons is considered to be oxidized. In the case of this reaction, Fe in the FeCl₃ is reduced into Fe by gaining 3 electrons - while Mg is oxidized to MgCl₂ because it donates 2 electrons.
If you check the moles of Mg, FeCl₃, Fe, and MgCl₂; you'll get to notice that the exchange of electron is balanced at 6.
Answer:
15.2L at STP
Explanation:
Given reaction expression;
CaCO₃ → CaO + CO₂
Number of moles of CaCO₃ = 0.68mol
Unknown:
Volume of CO₂ produced at STP = ?
Solution:
To solve this problem, we must first find the number of moles CO₂ produced,
1 mole of CaCO₃ will produce 1 mole of CO₂
0.68mole of CaCO₃ will produce 0.68mole of CO₂ at STP
Now;
1 mole of gas occupies a volume of 22.4L at STP;
0.68mole of CO₂ will then occupy 0.68 x 22.4 = 15.2L at STP
