Redox reaction and electrode potentials

2008-05-19 8:57 am
1.Why is it usual carry out reactions involving the manganate(VII) ion in acid solution?

2.Why is it customary to acidify potassium manganate(VII) with dilute sulphuric acid rather then with dilute hydrochloric acid?

3.Order of the tendency of the equation below
a. Cl2 + 2e- <---> 2Cl-
b. I2 + 2e- <---> I-
c. Fe3+ + e- <---> Fe2+
d. Br2 + 2e- <---> 2Br-
e. MnO4- + 8H+ +5e- <---> Mn2+ + 4H2O
f. S4O6(2-) + 2e- <---> 2S2O3(2-)

回答 (1)

2008-05-19 10:31 am
✔ 最佳答案
1.
In acidic solutions, the oxidizing power of MnO4-(aq) ion is strong, and it is reduced Mn2+(aq).
MnO4-(aq) + 8H+(aq) + 5e- ≒ Mn2+(aq) + 4H2O(l)
with standard reduction potential aaEo = +1.51 V

If other media, the oxidizing power of MnO4-(aq) ion becomes weaker, and the product is different. For example, in neutral solution:
MnO4-(aq) + 2H2O(l) + 3e- ≒ MnO2(s) + 4OH-(aq)
with standard reduction potential aaEo = +0.59 V

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2.
Consider the following two half reactions:
MnO4-(aq) + 8H+(aq) + 5e- ≒ Mn2+(aq) + 4H2O(l) aaEo = +1.51 V
Cl2(aq) + 2e- ≒ 2Cl-(aq) aaEo = +1.36 V

Consider the following reaction :
2MnO4-(aq) + 16H+(aq) + 10Cl-(aq) → 2Mn2+(aq) + 8H2O(l) + 5Cl2(g)
Standard e.m.f. = (+1.51) - (+1.36) = +0.15 V > 0
The reaction is feasible under standard condition.
In other words, MnO4-(aq, 1M) solution can oxidize HCl(aq, 1 M) under standard condition.

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3.
The greater the standard reduction potential, the higher is the tendency of the forward reaction (reduction) under standard condition, and the lower the tendency of the backward reaction (oxidation) under standard condition.

Standard reduction potentials:
(a) +1.36 V
(b) +0.53 V
(c) +0.77 V
(d) +1.07 V
(e) +1.51 V
(f) +0.26 V

Tendency of forward reaction: (e) > (a) > (d) > (c) > (b) > (f)
Tendency of backward reaction: (f) > (b) > (c) > (d) > (a) > (e)


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