Consider the following cell diagram Cr(s) | Cr^3+ (0.150 M) || Fe^2+ (????? M) C

Question

Consider the following cell diagram Cr(s) | Cr^3+ (0.150 M) || Fe^2+ (????? M) Consider the voltaic cell below. Label the ANODE compartment, the CATHODE compartment. and the SAIT BRIDGE. Labet the electrode that is Cr. label the electrode that b Fe. Indicate the direction of electron flow. Show the direction that the ions in the salt bridge will move. Show the direction that the ions in the cell compartments move. Which electrode will gain mass and which electrode will lose mass as the cell reaction proceeds? What is the overall reaction for this cell? What is the E_Cell for this What is the equilibrium constant for this reaction? You will/should obtain a really big number. Now that you hair K_BQ. what is Delta G for this reaction? If the potential of the cell. E_Cell was measured to be 0.283 volts, what was the concentration of the iron(II) ion? List one other suitable REDUCING agent and explain why it would be a suitable REDUCING agent for this cell. list one other suitable OX IIH ZING agent and explain why it would be a suitable OXIDIZING agent for this cell.

Explanation / Answer

Fe2+ (aq.) + 2e- ----------> Fe0 (s)        E0Fe2+/Fe = –0.44 V

Cr3+ (aq.) + 3e- -----------> Cr0 (s)       E0Cr3+/Cr = ­–0.74 V

As Standard Reduction potential of Cr is more –ve than that of Fe electrode.

Hence Cr will act as reductant and itself will get oxidized to Cr3+ (aq.)

Hence oxidation will occur at Cr electrode and reduction at Fe electrode

Hence,

At cathode, Fe2+ (aq.) + 2e- ----------> Fe0 (s)

At anode, Cr0 (s) --------------> Cr3+ (aq.) + 3e-

Balanced chemical equation is,

3 Fe2+ (aq.) + 2Cr0 (s) ----------> Fe0 (s) + 2Cr3+ (aq.)

Cr will go into the solution in Cr3+ form and Fe will get deposited on Fe electode.

Total of 6e transfer reaction i.e. n = 3

Equilibrium constant KEQ is given as,

KEQ = [Cr3+]2/[Fe2+]3 ……………….(1)

E0Cell = E0cathode – E0anode

E0Cell = E0Fe2+/Fe – E0Cr3+/Cr

E0Cell = (– 0.44) – (– 0.74)

E0Cell = +0.30V

Standard EMF of cell is +0.30V

=============== ================

At 298 K Equilibrium constant (KEQ) and E0Cell are related as,

E0Cell = (0.0591/n) ln (KEQ)

For given redox reaction we have,

E0Cell = +0.30 V, n = 6,

Let us put these values in eq. (2)

+0.30 = (0.0591/6) ln (KEQ)

0.30 = 0.00985 ln(KEQ)

ln(KEQ) = 0.30/0.00985

ln(KEQ) = 30.46

KEQ = e30.46

KEQ = 1.69 x 1013

Equilibrium constant is 1.69 x 1013

==============================

3) Equilibrium constant (K) and Gibbs energy change(0G) are related as,

0G = – RT ln(K)

0G = – 8.314 x 290x 30.46 …………………(ln(K) directly used)

0G = – 73440 J/mol

0G = –73.440 kJ/mol

Gibbs free energy change is -73.440 kJ/mol.

============================================

4) [Fe2+] = ?

Equilibrium constant (KEQ) and E0Cell are related as,

E0Cell = (0.0591/n) ln (KEQ)

With give E0Cell = 0.283 V let us calculate KEQ

0.283 = (0.0591/6) X ln(KEQ)

0.283 = 0.00985 ln(KEQ)

ln(KEQ) = 0.283/0.00985

ln(KEQ) = 28.73

KEQ = e28.73

KEQ = 3 x 1012

With this value of KEQ and [Cr3+] = 0.150 M in eq. (1)

KEQ = [Cr3+]2/[Fe2+]3

3 x 1012 = (0.150)2/[Fe2+]3

[Fe2+]3 = (0.150)2 /3 x 1012

[Fe2+]3 = 7.5 x 10-15

[Fe2+]3= 1.96 x 10-5 M

Concentration of Fe2+ is 1.96 x 10-5 M.

==========================================

5)

i) More –ve the standard reduction potential more is the tendency of metal to act as reductant and hence any electrode with Standard reduction potential more –ve than – 0.44V will act as reductant i.e. reducing agent.

e.g. Zn with standard reduction potential -0.76 V (It means we replace Cr electrode by Zn)

ii) And any other electrode with standard reduction potential less –ve than -0.74 V will act as oxidant i.e. oxidizing agent.

e.g. Cd with standard reduction potential -0.40 V (it means we replace Fe electrode by Cd)

==========================XXXXXXXXXXXXX==========================================

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.