Treatment of uranium (U) -contaminated groundwater has been accomplished by a bi

Question

Treatment of uranium (U) -contaminated groundwater has been accomplished by a bioremediation approach in which bacteria mediate the transformation of uranium for a soluble form (UO_2^2+) to a less soluble form (UO_2(s)). Fill in the blanks to balance the reaction of UO_2^2+ with lactate (C_3H_5O_3^-) 6 UO_2^2+ _ C_3H_5O_3^- + _H_2O = _UO_2(s) + _CO_2(aq) + _H^+ Determine the oxidation states for U in UO_2(s) and UO_2^2+ and C in C_3H_5O_3^- and CO_2(aq), respectively. What is the minimum amount of lactate (in mg/L) necessary for the complete reduction of 1 mg/L of UO_2^2+? The lactate is consumed in the aquifer following first-order kinetics with a rate constant of 0.7 d^-1. The acetate is injected directly upgradient of a uranium-contaminated zone that is 30.5 m long. In order to always have a concentration greater than 0.1 mg/L in the contaminated zone and to not add more lactate than is necessary, what should the lactate concentration be just after it is injected in the groundwater? The groundwater has a hydraulic gradient of 0.001. The aquifer is coarse sand with a hydraulic conductivity of 2000 m/d and a porosity of 0.35. Assume the groundwater moves like a plug flow and the lactate transport is not retarded.

Explanation / Answer

a+b) We will consider the change in oxidation numbers of the reactants to determine the balanced chemical equation.

UO22+ + 2 e- --------> UO2 (reduction)

2 C3H5O3- + 6 H2O --------> 6 CO2 + 8 e- + 22 H+ (oxidation)

To balance the two reactions, multiply the reduction half by 4 and the oxidation half by 1 and add.

4 UO22+ + 2 C3H5O3- + 6 H2O -------> 4 UO2 + 6 CO2 + 22 H+ (balanced) (ans)

The oxidation state of U in UO22+ and UO2 can be determined as below:

Let the oxidation number of U in the two compounds be x1 and x2. Therefore,

x1 + 2*(-2) = +2

===> x1 – 4 = +2

===> x1 = +2 + (+4) = +6

x2 +2*(-2) = 0

===> x2 – 4 = 0

===> x2 = +4

The oxidation numbers of U in UO22+ and UO2 are +6 and +4 respectively (ans).

Let the oxidation numbers of C in C3H5O3- and CO2 be y1 and y2 respectively.

3*y1 + 5*(+1) + 3*(-2) = -1

===> 3y1 + 5 – 6 = -1

===> 3y1 -1 = -1

===> 3y1 = -1 + (+1) = 0

y2 + 2*(-2) = 0

===> y2 – 4 = 0

===> y2 = +4

The oxidation numbers of carbon in C3H5O3- and CO2 are 0 and +4 respectively (ans).

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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