Chemistry, asked by Hilal5969, 1 year ago

Coni nanoparticles/cnt composite as effective anode for direct urea fuel cells

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Answered by ravi9848267328
0

Answer:

Explanation:

1. INTRODUCTION

Besides the human and animal urines, fertilizers industries discharge huge amount of ureacontaminated wastewaters. The urea fertilizer manufacturing plants discharge 0.75 m3 wastewater

containing around 1wt% urea per ton fertilizer produced [1]. Environmentally, the urea-polluted water

should be treated to avoid formation of the ammonia gas from urea hydrolysis [2].

NH2CONH2 + H2O 2NH3 + CO2 (1)

Besides the air pollution which may result in falling the acidic rains, the produced ammonium

can create other environmental problems. The dissolved ammonia is biologically oxidized in water by

two kinds of microorganisms (Nitrobacter and Nitrosomonas ) to nitrate (NO3) in a process called  

Int. J. Electrochem. Sci., Vol. 13, 2018 4694

nitrification[3]. The main problem is producing a dangerous unstable compound as an intermediate;

nitrogen dioxide. This process takes place in the oxygen-containing media such as open water surfaces

so the nitrate ion can be reduced to form different kinds of nitrogenous gases. Beside the

aforementioned problems, domoic acid which is deadly toxin can be produced by some kinds of ocean

algae in the urea-polluted media [4].

Owing considerable amount of hydrogen (6.67 wt%) drew the attention of the researchers to

exploit the urea-containing wastewater as a good source of hydrogen gas. Therefore, electrolysis has

been proposed as a treatment process for the urea-containing wastewater and simultaneously producing

H2 gas due to occurring some oxidation reactions which can be explained electrochemically as follow

[5-7]:

Anode: CO(NH2)2 + 6OH¯ N2 + 5H2O + CO2 + 6e E

0

= - 0.746 V (2)

Cathode: 6H2O + 6e 3H2 + 6OH¯ E

0

= - 0.829 V (3)

Overall: CO(NH2)2 + H2O N2 + 3H2 + CO2 E

0

= - 0.083 V (4)

As it might be concluded from the standard potentials, the electrolysis process needs small

energy however, practically, due to the high overpotentials, the real electrode potentials are very far

from the theoretical ones especially the anode potential even with the precious metals anodes[8].

Therefore, the electrolysis process needs high energy and it is economically not preferable.

Developing a proper anode material can lead to operate direct urea fuel cells (DUFC) in which

the urea is oxidized and simultaneously the wastewater will be treated. In the DUFC, the electrodes

reactions will be in these forms [9].

Cathode: 2H2O+O2+4e 4OH¯ E0

= + 0.40 V (5)

Anode: CO(NH2)2+6OH¯ N2+5H2O+CO2+6e E

0

= - 0.746 V (6)

Overall: 2CO(NH2)2+3O2 2N2+4H2O+2CO2 E

0

= + 1.146 V (7)

Answered by kashish762001
0

Answer: plz mark this answer as the brainliest

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