(1) Au+(aq) + e- Au(s) E° = 1.69 V (2) N2O(g) + 2 H+(aq) + 2 e- N2(g) + H2O(l) E

Question

(1) Au+(aq) + e- Au(s) E° = 1.69 V
(2) N2O(g) + 2 H+(aq) + 2 e- N2(g) + H2O(l) E° = 1.77 V
(3) Cr3+(aq) + 3 e- Cr(s) E° = -0.74 V



** Please help me fix the ones with the red X next to them *****

Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizing and reducing agents for each reaction. (Type your answer using the format [NH4]+ for NH4, [Mg]2+ for Mg2+, and CO2 for CO2. Don't specify physical states. Use the lowest possible coefficients.) (1) Au+(aq) + e- Au(s) (2) N20(g) + 2 H+(aq) + 2 e- N2(g) (3) Cr3+(aq) + 3 e-Cr(s) E" = 1.69 V H20(,)E" = 1.77 V E0.74 V a) 3 X (aq) 2 Cr(s) 3 N2 (g)3 E° cell 2.51 oxidizing agents H2O reducing agents [N20] (b) 3 Au+(aq) 1 Cr (s)1 E° cell 2.43 oxidizing agents Cr reducing agents AU N2O (g) 1 (g)2 H2O (aq) +2 N2 E° cell 0.08 oxidizing agents N20 reducing agents N2

Explanation / Answer

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

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