Relative Half-Cell Potentials Assuming standard conditions, answer the following
ID: 1056743 • Letter: R
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.)
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)?
Explanation / Answer
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.