Optional Homework 6 for practice only!!!
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1 Name: Date: Optional Homework 6 for practice only!!! 1.) Write the oxidation number for each atom in the molecules below: a. Gallium sulfide Ga 2 S 3 Ga +3 S -2 b. Chromium (III) oxide Cr 2 O 3 Cr +3 O -2 c. Potassium phosphite K 3 PO 3 K +1 O -2 P +3 d. Calcium phosphate Ca 3 (PO 4 ) 2 Ca +2 O -2 P +5 e. Aluminum carbonate Al 2 (CO 3 ) 3 Al +3 O -2 C +4 2.) Write the half-reactions for each of the following chemical reaction including the electrons and indicate exactly which species is oxidized, which is reduced, which is the oxidizing agent and which is the reducing agent: they are not balanced please do no worry/balance them!! a. PbO (s) + CO (g) Pb (s) + CO 2 (g) Pb +2 O -2 C +2 O -2 Pb 0 C +4 O -2 Pb e -1 Pb 0 C +2 C e -1 Pb +2 is reduced, therefore it is the oxidizing agent C +2 is oxidized, therefore it is the reducing agent b. Fe (s) + Cl 2 (g) FeCl 3 (s) Fe 0 Cl 2 0 Fe +3 Cl -1 Fe 0 Fe e -1 Cl 2 0 2Cl e -1 Cl 2 0 is reduced, therefore it is the oxidizing agent Fe 0 is oxidized, therefore it is the reducing agent
2 c. CO (g) + I 2 O 5 (s) I 2 (s) + CO 2 (g) C +2 O -2 O -2 I +5 I 2 0 C +4 O -2 2I e -1 0 I 2 C +2 C e -1 I +5 is reduced, therefore it is the oxidizing agent C +2 is oxidized, therefore it is the reducing agent d. Cu (s) + HNO 3 (aq) Cu(NO 3 ) 2 (aq) + NO 2 (g) + H 2 O (l) Cu 0 H +1 N +5 O -2 Cu +2 N +5 O -2 N +4 O -2 H +1 O -2 N e -1 N +4 Cu 0 Cu e -1 N +5 is reduced, therefore it is the oxidizing agent Cu 0 is oxidized, therefore it is the reducing agent 1.) Balance the following chemical reactions as indicated identify the reducing and oxidizing species: (insert spaces as needed or perform on a separate sheet of paper) a. SO 3-2 (aq) + Cl 2 (g) SO 4-2 (aq) + Cl -1 (aq) [basic] S +4 O -2 Cl 0 S +6 O -2 Cl -1 SO 3-2 is oxidized to SO 4-2 therefore SO 3-2 is the reducing agent Cl 2 is reduced to Cl -1 therefore Cl 2 is the oxidizing agent SO -2 3 SO -2 4 Cl 2 Cl -1 SO H 2 O SO -2 4 Cl 2 2Cl -1 SO H 2 O SO H +1 Cl 2 + 2e -1 2Cl -1 SO H 2 O SO H e -1 SO H 2 O SO H e -1 Cl 2 + 2e -1 2Cl -1 SO H 2 O + Cl 2 SO Cl H +1 2OH -1 + SO H 2 O + Cl 2 SO Cl H OH -1 2OH -1 + SO H 2 O + Cl 2 SO Cl H 2 O 2OH -1 + SO Cl 2 SO Cl -1 + H 2 O
3 2.) Balance the following chemical reactions as indicated identify the reducing and oxidizing species: (insert spaces as needed or perform on a separate sheet of paper) a. MnO 4-1 (aq) + H 2 O 2 (aq) Mn +2 (aq) + O 2 (g) [acidic] Mn +7 or MnO4-1 is being reduced, therefore it is the oxidizing agent O -1 or H2O2 is being oxidized, therefore it is the reducing agent (we can only have 1 species being reduced and 1 species being oxidized, since MnO4-1 is tied to the Mn +2 that leaves H2O2 to be tied to O2) MnO -1 4 Mn +2 H 2 O 2 O 2 MnO -1 4 Mn H 2 O H 2 O 2 O H +1 8 H +1 + MnO -1 4 Mn H 2 O H 2 O 2 O 2 + 2H e -1 2(5e H +1 + MnO 4-1 Mn H 2 O) 5(H 2 O 2 O 2 + 2H e -1 ) 10e H MnO 4-1 2Mn H 2 O 5H 2 O 2 5O H e -1 16H H 2 O 2 + 2MnO 4-1 2Mn O 2 + 8H 2 O + 10H +1 6H H 2 O 2 + 2MnO 4-1 2Mn O 2 + 8H 2 O
4 3.) Consider the following general voltaic cell: voltmeter A B E C D Identify the anode, cathode, salt bridge, electrode at which the electrons accumulate (negatively charged electrode), the electrode at which the electrons are consumed (the positively charged electrode), the electrode that would gain mass as the cell operates (assuming that the metal plates out of solution), and the electrode that would lose mass as the cell operates E = salt bridge A = anode: the anode is negatively charged. Oxidation is happening at the anode so electrons are being produced there and are in excess. Since oxidation is happening there, the electrode M is turning into M +x ions so the electrode is losing mass B = cathode: the cathode is positively charged. Reduction is happening at the cathode so electrons are being consumed there and the area is deficient. Since reduction is occurring there, the ions in solution are turning into solid metal, plating onto the electrode and thus the electrode is gaining mass C and D are both electrolytic solutions 4.) Using the activity series, predict which of the following reactions will occur spontaneously AS WRITTEN: remember that with single replacement reactions you can form H 2 gas b. Cu (s) + Zn(NO 3 ) 2(aq) NR Zn +2 prefers to stay as the ion. Cu cannot reduce it c. 2Al (s) + 3Pb(NO 3 ) 2(aq) yes 2Al(NO 3 ) 3 + 3Pb (s) Al prefers to be oxidized d. Al (s) + 3HCl (aq) yes AlCl 3 + 3/2 H 2 (g) Al prefers to be oxidized e. Mg (s) + 2HCl (aq) yes MgCl 2 + H 2 Mg prefers to be oxidized
5 f. Fe (s) + CuSO 4(aq) yes iron prefers to be oxidized (iron II or iron III could be a product) g. Fe (s) + AgC 2 H 3 O 2(aq) yes, iron again prefers to be oxidized h. Fe (s) + NaBr (aq) NR sodium prefers to be the oxidized species (prefers to be the ion) i. AgNO 3(aq) + Cu (s) yes copper prefers to be oxidized 5.) CALCULATE E o for the oxidation half-reaction Co (s) Co +2 (aq) + 2e - given the cell voltage for the following voltaic cell (Platinum does not participate in the overall chemical reaction) Co (s) Co +2 (1 M) Ce +4 (1 M), Ce +3 (1 M) Pt (s) E o cell = V E reduction = 1.44 V E cell o = E red + E ox V = 1.44 V + E ox E ox = 0.45 V 6.) Balance the following given half-reactions and determine the overall balanced redox reaction and the E o cell for the overall reaction: O 2 (g) H 2 O (l) and I -1 (aq) I 2 (s) 4e - + 4H +1 + O 2 2H 2 O 2I -1 I 2 + 2e -1 E reduction = V E reduction = 0.54 V but it s written as oxidation E oxidation = V 4e - + 4H +1 + O 2 H 2 O (2I -1 I 2 + 2e -1 )2 4e H +1 + O 2 H 2 O 4I -1 2I 2 + 4e -1 4H I -1 + O 2 H 2 O + 2I 2 E cell = E red + E ox = (-0.54V) = 0.69 V
6 7.) Determine E o cell values for the following redox reaction: 2Al (s) + 3Cu +2 (aq) Al +3 (aq) + 3Cu (s) Al Al e -1 Cu e -1 Cu table Al e -1 Al E red = V E red = 0.34 V but this is written as oxidation so change the sign to V Al Al e -1 E ox = V Cu e -1 Cu E red = 0.34 V E cell o = V 8.) Will copper metal displace silver ion from aqueous solution? That is, does the reaction occur spontaneously from left to right or right to left. Prove this mathematically by calculating E o cell. Cu (s) + 2Ag +1 (aq) Cu Ag (s) Cu Cu e -1 (from the table of reduction potentials we see the reduction value as 0.34 V, since this is written as oxidation, we change the sign) E oxidation = V 2Ag e -1 2Ag (s) E reduction = 0.80 V Cu Cu e -1 E ox = V 2Ag e -1 2Ag (s) E red = V Cu(s) + 2Ag +1 (aq) Cu Ag(s) Ecell o = V the reaction IS SPONTANEOUS as written since E is positive 9.) Write the electron configurations DO NOT USE THE NOBLE GAS CONFIGURATION!! for the following: a. Fe 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 b. Zn 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10
7 10.) Write a set of quantum numbers (n, l, m l, and m s ) for the 5 th electron, the 10 th electron, the 12 th electron, the 20 th electron, the 27 th electron, and the 30 th electron for zinc. (HINT!! Orbital box diagrams might help!) s 2s 2p 3s 3p s 3d 5 th e: n = 2 l = 1 m l = -1 m s = + ½ 10 th e: n = 2 l = 1 m l = 1 m s = - ½ 12 th e: n = 3 l = 0 m l = 0 m s = - ½ 20 th e: n = 4 l = 0 m l = 0 m s = - ½ 27 th e: n = 3 l = 2 m l = -1 m s = - ½ 30 th e: n = 3 l = 2 m l = +2 m s = - ½ 11.) Write the electron configurations DO NOT USE THE NOBLE GAS CONFIGURATION!! for the following a. K +1 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 b. Cr +3 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 3 c. Cr +6 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 0
8 12.) Which of the following quantum number sets are allowed? If there is an error identify AND correct it by changing the l number only The n number is the principal quantum number. It can be any number greater than 1. The l value tells you what shape orbital the electron is in. These are FIXED values. If l = 0 it represents an s orbital. All values of n can have an s orbital If l = 1 it represents a p orbital. Values of n =2 and higher can have an p orbital If l = 2 it represents a d orbital. Values of n =3 and higher can have an d orbital If l = 32 it represents a f orbital. Values of n =4 and higher can have an f orbital And so on... after the f, we go alphabetically with the g orbital, the h, the i... Essentially, the MAXIMUM l value is n-1. You cannot have an l value = n. All other l values that are lower than n-1 are acceptable. The m l quantum numbers states specifically what orbital the electron is in. Remember there are 3 p orbitals, the p x the p y and the p z. Therefore there are 3 numbers that correspond to those orbitals. Direct assignment does not exist. This is your freedom of choice quantum number. The values of m l run from - l to + l. Therefore, if l = 0, the only possible m l value is 0. But if l = 1, then m l runs from Which means -1, 0, and 1 are all possible acceptable values. This should make sense because if l = 1 we are talking about the p shaped orbital. There are 2 p orbitals, and we now have 3 possible numerical values. a. n= 3, l = 2, m l = 2 VALID: l is less than n and the m l value is within the acceptable range of l to + l b. n=6, l =5, m l = -4 VALID: The l value is less than the n value and the m l is in the acceptable range
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