Question

Relative Half-Cell Potentials

Assuming standard conditions, answer the following questions.
(Use the table of Standard Reduction Potentials for common Half-reactions from your text.
If hydrogen is one of the reagents, assume acidic solution.)

cf Table 11.1 on p 480 of Zumdahl 7th ed.

yes no  Is H₂(g) capable of reducing Ag⁺(aq)?
yes no  Is Fe²⁺(aq) capable of reducing Cr³⁺(aq) to Cr metal?
yes no  Is Fe²⁺(aq) capable of oxidizing Cr metal?
yes no  Is Fe³⁺(aq) capable of oxidizing Sn metal to Sn²⁺(aq)?
yes no  Is Fe²⁺(aq) capable of reducing Cr³⁺(aq) to Cr²⁺(aq)?
yes no  Is VO₂⁺(aq) capable of oxidizing Fe²⁺(aq)?

Answer 1

Oxidizing agent has a tendency to accepts electrons and reducing agent accepts electrons.

the reactions are
Ag+(aq) + e- → Ag(s) Eo=0.80 -1 and 2H+ +2e- --------> H2(g) Eo=0 (2)

Multiplying Eq.1 with and 2 reversing gives 2Ag(s) -------> 2Ag+ +2e- Eo=-0.80 (1A)

and 2H+ + 2e- ---------> H2 Eo= 0 (2)

addition gives 2Ag+ 2H+ -----> 2Ag+ + H2 Eo= -0.80

higher reduction potential, easier reduced or more capable of oxidizing and lower reduction potential, easier oxidized or more capable of reducing .

2.Fe2+(aq) + 2e- → Fe(s) Eo=0.44V and Cr3+ (aq) + 3e- → Cr (aq) Eo=-0.74V Hence Fe+2 is not capable of reducing Cr+3 to Cr(s).

3Fe+2 is not capable of oxdizing Cr metal.

3. Fe3+(aq) + 3e- → Fe(s) E0=-0.04V
Sn2+(aq) + 2e- → Sn(s) Eo= -0.14

So Fe+2 is capable of o oxdizing.

5.  Fe2+(aq) + 2e- → Fe(s) E0=-0.44V and Sn2+(aq) + 2e- → Sn(s) Eo= -0.14V and Hence FFe+2 is capable

6. VO2+ is capable to oxdizie Fe+2.