Question

Part 1.)

Using data found in Appendix E of your textbook calculate the nonstandard emf for each of the following reactions if the concentration of each of the ions in these reactions is 0.0002 molar and everything else is standard (use 298 K for the temperature, R = 8.314 J/mol-K, and F = 96,485 C/mol):

(a) 2 Hg2+(aq) + 1 Sn(s) --> 1 Hg22+(aq) + 1 Sn2+(aq)	E = __ V
(b) 1 F2(g) + 1 Mg(s) --> 2 F-(aq) + 1 Mg2+(aq)	E = __ V
(c) 2 Cr3+(aq) + 3 Co(s) --> 2 Cr(s) + 3 Co2+(aq)	E = __ V
(d) 1 Fe2+(aq) + 2 Co2+(aq) --> 1 Fe(s) + 2 Co3+(aq)	E = __ V

Part 2.)

Choose all of the statements from below which are true about strengths of oxidants and reductants.

1.The strengths of oxidants and reductants on opposite sides of redox equations correlate oppositely.
2. For a strong reductant E^o_red should be negative.
3. Reducing agents are found on the right-hand side of reduction half reactions.
4. For a strong oxidant E^o_red should be positive.
5. Oxidizing agents are found on the right-hand side of reduction half reactions.

Answer 1

a) When you are dealing with this problems think of Nernst equation

E = E⁰ + (RT/nF) * ln (a_ox/a_red)

Putting all the values T = 298 K R = 8.314 JK^-1mol^-1 equation simplifies

E = E⁰ + (0.059/n) * log (a_ox/a_red)

Now this is for half cell E= potential E⁰ = Std. potential a = respective activity(you can replace by concentration also)

For a full cell this equation becomes

$\Delta$E = $\Delta$ E⁰ + (0.059/n) * log (a_ox/a_red) where $\Delta$ E⁰ = E⁰(RHE) - E⁰(LHE) RHE and LHE are right hand and left hand electrode.

if the concentration of each of the ions in these reactions is 0.0002 molar you actually don't need to worry about hte concentration part , it will cancel out . So the equation will become just

$\Delta$E = $\Delta$ E⁰ = E⁰(RHE) - E⁰(LHE)

Take std values from your books and calculate them

RHE is always that one whose reduction potential is higher.

b) 1.The strengths of oxidants and reductants on opposite sides of redox equations correlate oppositely.(yes)
2. For a strong reductant E^o_red should be negative.(Reductant donate electrons , so more they easily doante electron, more it becomes spontaneous, so it is correct,E^o_red goes negative .)
3. Reducing agents are found on the right-hand side of reduction half reactions.(wrong they are the one who donate electrons found in left side)
4. For a strong oxidant E^o_red should be positive.(Oxidant accept electrons , so more they easily accept electron, more it becomes spontaneous, so it is correct.)
5. Oxidizing agents are found on the right-hand side of reduction half reactions.( correct they are the one who accept electrons found in left side)