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MAVERICK [17]
3 years ago
11

Balance the following redox equation in acidic solution using the smallest integers possible and select the correct coefficient

for the H+(aq) ion.Cr2O72–(aq) + Sn2+(aq) → Cr3+(aq) + Sn4+(aq)(A) 1 (no coefficient written)(B) 2(C) 3(D) 4(E) More than 4
Chemistry
1 answer:
Evgen [1.6K]3 years ago
4 0

Answer:

The balanced redox is:

14 H⁺ + Cr₂O₇²⁻ + 3Sn²⁺ → 3Sn4⁺  + 2Cr³⁺  + 7H₂O

So the coefficient for the H⁺ is greater than 4 (option E)

Explanation:

This is the redox reaction:

Cr₂O₇²⁻ (aq) + Sn²⁺ (aq)  →  Cr³⁺ (aq) + Sn⁴⁺(aq)

First of all, we must determine the half reactions:

In dycromate, Cr acts with +6 in the oxidation state → Cr cation has +3 in product side - Oxidation state, has decreased so this is the reduction.

In reactant side Sn cation acts with +2 → In product side Sn acts with +4

The oxidation state has increased, so this is the oxidation.

Cr₂O₇²⁻  → Cr³⁺

We have to add 2, to Cr in reactant side, and as we are in adicid medium we add water in the opposite side of oxygen. The same amount of oxgen, that we have.

Cr₂O₇²⁻  → 2Cr³⁺ + 7H₂O

Finally, as we have 14 H in product side, we must add 14 H⁺ to the reactant side. Cr+⁶ in dycromate to change to Cr³⁺, gained 3 e⁻, but we have 2 Cr, so in total the Cr gained 6e⁻. The balanced half reaction is:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺  + 7H₂O

Sn²⁺ to change the oxidation state, to +4 had to release electrons:

Sn²⁺ →  Sn4⁺  + 2e⁻

The electrons are unbalanced, so we must to multiply the half reactions:

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺  + 7H₂O) x1

(Sn²⁺ →  Sn4⁺  + 2e⁻ ) x3

And we sum both:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 3Sn²⁺ → 3Sn4⁺  + 6e⁻  + 2Cr³⁺  + 7H₂O

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7 0
3 years ago
14. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km
emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

v = \frac{vmax[S]}{Km + [S]}

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

t = 27 min

7 0
3 years ago
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