Answer the following questions r to following electrochemical ce elated 2NO(g) H
ID: 499154 • Letter: A
Question
Answer the following questions r to following electrochemical ce elated 2NO(g) H20 (l) 2e N20 g) 20H (aq) EO 0.760 V Bro3 (aq) 3H20 (l) 6e Br (aq) 60H (aq) EO 0.610 V 1. Answer the following questions under standard conditions. (a) What is E cel (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V) (b) which one of the following chemical reactions describes the process occurring in the electrochemical cell? O 3N20 (g) Bro3 (aq) 6NO(g) Br (aq) O ON 20 (g) 3Bro3-(aq) 12NO(g) 2Br (aq) O 12NO(g) 2Br (aq) ON 20 (g) 3Bro3-(aq) O 6NO(g) Br (aq) 3N20 (g) Bro3 (aq) (c) What is AG (in kJ/mol)? Report your answer to three significant figures in scientific notation (i.e., 1.23e2 kJ/mol).Explanation / Answer
1. For the given electrochemical cell,
(a) Eocell = Ecathode - Eanode
= 0.760 - 0.610
= 0.150 V
(b) The correct cell reaction would be,
6NO(g) + Br-(aq) --> 3N2O(g) + BrO3-(aq)
(c) Using,
dGo = -nFEo
= -6 x 96500 x 0.150/1000
= -86.850 kJ/mol
2. For the given cell,
(a) Ecell = Eocell - RT/nF lnK
Feed the given values,
Ecell = 0.150 - (8.314 x 344.7/6 x 96500) ln[(0.622)^3.(4.1)/(0.840)^6.(6.99 x 10^-5)]
= 0.0975 V
(b) the cell reaction would be,
6NO(g) + Br-(aq) --> 3N2O(g) + BrO3-(aq)
(c) dG = -RTlnK
= -8.314 x 344.7 ln[(0.622)^3.(4.1)/(0.840)^6.(6.99 x 10^-5)]/1000
= -30.4 kJ/mol
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