Use the information, below, to answer the next two (2) questions. Half-reaction (V) Cl2 + 2e + 2 C1° 1.36 Ag+ + e - Ag...
Table 20.1 Half Reaction E°(V). F2 (g) + 2e →2F (aq) +2.87 Cl2 (g) + 2e → 2CV (aq) +1.359 Br2 (1) + 2e → 2Br (aq) +1.065 O2 (g) + 4H+ (aq) + 4e → 2H20 (1)+1.23 Agt te → Ag (s) +0.799 Fe3+ (aq) + € → Fe2+ (aq) +0.771 12 (s) + 2e → 21+ (aq) +0.536 Cu2+ + 2e → Cu(s) +0.34 2H+ + 2e → H2 (g) Pb2+ + 2e → Pb (s) -0.126 Ni2+...
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + € → Fe2+ +0.77 v +0.34 V -0.13 V Cu2+ +2e → Cu Pb2+ + 2e - → Ib Ni2+ + 2e → Ni Cd2+ +2e → Cd -0.25 V -0.40 V Fe2+ + 2e → Fe -0.44 V Zn2+ + 2e → Zn -0.76 V Al3+ +3e → AI - 1.66 V Consider an electrochemical cell constructed from the following half cells, linked by...
19 20 Question 16 Half-cell Potentials: Half Reaction: E' value + 0.80 V +0.77 V Agte → AS Fe3+ + + Fe2+ Cu2+ 2e → Cu Pb2+ + 2e → Pb +0.34 V -0.13 V Ni2+ + 2e → NI -0.25 V - 0.40 V Cd2+ +2e → ca Fe2+ + 2e → Fe Zn2+ + 2e → Zn -0.44 V - 0.76 V A13+ +3 → AI - 1.66 V Consider an electrochemical cell constructed from the following half...
Consider the following half-reactions: Half-reaction E° (V) Ag+(aq) +e → Ag(s) 0.799V Cu2+(aq) + 2e → Cu(s) 0.337V Mg2+(aq) + 2e —— Mg(s) -2.370V The strongest oxidizing agent is: enter symbol The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will Mg(s) oxidize Ag(s) to Ag+(aq)? V Which species can be oxidized by Cu2+(aq)? If none enter none. Consider the following half-reactions: Half-reaction E° (V) 2Br (aq) 1.080V Br2(1) + 2e — Ni2+(aq) +...
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e →2F 0+ 2H + 2e →H,02 0.68 Ag* + e → Ag Cu* + e → Cu 0.52 Co? + e Co? O2 + 2H,0 + 4e → 40H 0.40 H,O, + 2H + 2e → 2H,0 Cu2+ + 2e → Cu Cet+e → Ces Cu2+ + e → Cut 0.16 PbO, + 4H+ +50,2 +2e → PbSO, + 2H,0 2H* + 2e →...
Fill in the Blanks Ered (V) 0.68 0.52 0.40 0.34 0.16 0 Half Reaction Ered (V) Half Reaction F2+2e →2F 2.87 02 + 2H+ 2e →H,O, Ag?+ e → Ag 1.99 Cute → Cu Code → CO2 1.82 O2 + 2H,0 + 4e → 40H H,02 + 2H+ + 2e → 2H,0 1.78 Cu2+ + 2e → Cu Ce+ + → Ce+ 1.70 Cu2+ e → Cu PbO, + 4H* +502 +2e → PbSO4 + 2H,0 1.69 2H*+2e → H2...
2.87 Ered® (V) 0.68 0.52 0.40 0.34 0.16 Half Reaction F,+ 2e →2F Ag* + e → Ag Co3 + e + CO2- H2O2 + 2H+ + 2e → 2H,0 Ce4+ + e → Ce+ PbO, + 4H+ + SO42- + 2e → PbSO, + 2H,0 Mno, + 4H+ + 3e → MnO2 + 2H,0 2e + 2H+ + 10, → 103 + H2O Mn0, +8H+ + 5e → Mn2+ + 4H,0 Aul+ + 3e → Au Cl2 + 2e...
A. Half-reaction E° (V) Cl2(g) + 2e-2Cl-(aq) 1.360V Sn2+(aq) + 2e-Sn(s) -0.140V Cr3+(aq) + 3e-Cr(s) -0.740V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Cl2(g) oxidize Cr(s) to Cr3+(aq)? (6) Which species can be oxidized by Sn2+(aq)? If none, leave box blank. B. Half-reaction E° (V) Cl2(g) + 2e-2Cl-(aq) 1.360V Cu2+(aq) + 2e-Cu(s) 0.337V Mn2+(aq) + 2e-Mn(s) -1.180V (1) The...
Please combine the Mn electrode half reaction with the Ag
electrode half reaction and write the complete redox reaction using
date from table 8.1. Indicate which is the oxidizing and which is
the reducing agent.
b. Calculate the EMF when the reaction quotient (Q) =
]Mn2+]/[Ag+]^2 = 10^-5
Reducing agent Half-reaction E0 (volts) -2.93 2.87 Ca Na Mg Mn Zn Fe Ni Pb Ca Ca2++2e Na → Na++e- -2.36 Mn Mn++2e Zn Zn2++2e Fe → Fe2++ 2e- NiNi2++2e- -0.76 -0.47...
a.
Cl2
b.
F2
c.
Br2
d.
I2
e.
All of the halogens have equal strength as oxidizing agents.
E°(V +2.87 +1.359 +1.065 +1.23 +0.799 Table 20.1 Half Reaction F2 (8) 2e → 2F- (ag) Cl2(8) 2e → 2Cl- (aq) Br21) 2e → 2Br" (ag) 02 (elut 4H+ (ag) + 4e + 2H20 (1) Agte → Ag (5) Fe3+ (aq) → Fe2+ (aq) 12 (5) + 2e – 21 (ag) Cu2+ + 2e - Cu (s) 2H+ + 2e →...