Answer:
As we move from top to bottom in the table, the ability to reduce others increases and on moving from bottom to top ability to oxidize others increases.
The element on the top of the table will be considered as strongest oxidizing agent and element in the bottom of the table will be considered as strongest reducing agent.
So, from the table, the weakest oxidizing agent belonging to halogens will be with Eo value greater than zero and smaller than 2.87V.
The increasing order of halogen's from weakest to strongest will be:

So, from this we can say that iodine is a weakest oxidizing agent belonging to the halogen family.
Table 2012 HalfReaction EM Fight 2 – 2 (aq) Cl, ig) + 2er -+ 20- (aq)...
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+...
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 →...
Standard Electrode Potentials at 25?C Reduction Half-Reaction E?(V) F2(g)+2e? ?2F?(aq) 2.87 Au3+(aq)+3e? ?Au(s) 1.50 Cl2(g)+2e? ?2Cl?(aq) 1.36 O2(g)+4H+(aq)+4e? ?2H2O(l) 1.23 Br2(l)+2e? ?2Br?(aq) 1.09 NO3?(aq)+4H+(aq)+3e? ?NO(g)+2H2O(l) 0.96 Ag+(aq)+e? ?Ag(s) 0.80 I2(s)+2e? ?2I?(aq) 0.54 Cu2+(aq)+2e? ?Cu(s) 0.16 2H+(aq)+2e? ?H2(g) 0 Cr3+(aq)+3e? ?Cr(s) -0.73 2H2O(l)+2e? ?H2(g)+2OH?(aq) -0.83 Mn2+(aq)+2e? ?Mn(s) -1.18 How can the table be used to predict whether or not a metal will dissolve in HCl? In HNO3? Drag the terms on the left to the appropriate blanks on the right to...
Using standard reduction potential in aqueous solutions at 25c Table, which substance is most likely to be oxidised by O2 (g) in acidic aqueous solution? Select one: a. Br2 (l) b. Br- (aq) c. Ni2+ (aq) d. Ag (s) e. Cu2+ (aq) Cathode (Reduction) Half-Reaction Standard Potential E° (volts) Li+(aq) + e- -> Li(s) -3.04 K+(aq) + e- -> K(s) -2.92 Ca2+(aq) + 2e- -> Ca(s) -2.76 Na+(aq) + e- -> Na(s) -2.71 Mg2+(aq) + 2e- -> Mg(s) -2.38 Al3+(aq)...
1. How do I read the half reaction table?
2. If im asked for the best reducing agent from Cu+, Ag+, F2, and
Fe3+, where do I look first in the table? before the arrow or after
the arrow?
3. Sometimes a value that has originally a positive (V) from the
table it will have the negative sign in a homework problem, and
viceversa, so the question is, how do I use the positive and
negative signs in respect to...
Find the best reducing agent from Cu+, Ag+ F2 and Fe3+
#1. In the reduction table i can see several repeated values of
Fe3+ one is equal to 0.77v and the second one is equal to -0.036v
so, which one do I choose? Please explain.
#2.If I'm asked to find the best oxidation agent, from the
values already provided (Cu+, Ag+ F2 and Fe3+) which one would it
be? and how would I decide from repeated values, like in #1,...
Sodium metal is prepared by electrolysis of sodium
chloride to the metal and chlorine gas. Using the
reduction potentials attached, calculate E° and ΔG° for the overall
reaction.
TABLE 12.2 Standard Reduction Potentials at 25°C Half-reaction Eº, V Increasing strength as oxidizing agent Lit(aq) + e Li(s) K+(aq) + e → KS) Ba2+ (aq) + 2e → Ba(s) Sr2+(aq) + 2e → Sr(s) Ca2+(aq) + 2e → Ca(s) Na*(aq) + e Na(s) Mg2+ (aq) + 2e → Mg(s) Be2+(aq) + 2e...
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.774 M and [Sn2 ] = 0.0190 M. Standard reduction potentials can be found here. Reduction Half-Reaction Standard Potential Ered° (V) F2(g) + 2e– → 2F–(aq) +2.87 O3(g) + 2H3O+(aq) + 2e– → O2(g) + 3H2O(l) +2.076 Co3+(aq) + e– → Co2+(aq) +1.92 H2O2(aq) + 2H3O+(aq) + 2e– → 2H2O(l) +1.776 N2O(g) + 2H3O+(aq) + 2e– → N2(g) + 3H2O(l) +1.766 Ce4+(aq) + e– → Ce3+(aq)...