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Use the following half-reactions to write three spontaneous reactions, calculateE∘cellfor each reaction, and rank the strengths of the oxidizing and reducing agents:

(1)Au+(aq) +e-nAu(s)E0= 1.69V

(2)N2O(g) + 2H+(aq) + 2e-nN2(g) +H2O(l)

(3)Cr3+(aq)+3e-nE∘=1.77VE∘=-0.74V

Short Answer

Expert verified

The reaction to be spontaneous to be combine and reversed solution and the cancel common terms of oxidising agents and reducing agent.

Step by step solution

01

 Step 1:  To Write three spontaneous reactions from the given half-reactions

(1)Au+(aq)+e-→Au(s)E∘=1.69VAu+(aq)+e-→Au(s)E∘=1.69V

(2)N2O(g)+2H++2e-→N2(g)+H2O(I)E∘=1.77V

(3)Cr3+(aq)+3e-→Cr(s)E∘=-0.74V

For a reaction to be spontaneous,E∘cell>0 (a) Combinerole="math" localid="1663760516388" 3×(1)and reversed of (3):

3Au+(aq)+3e-+Cr(s)→3Au(s)+Cr3+(aq)+3e-

Cancel common terms to obtain the spontaneous reaction:

3Au+(aq) + Cr(s)→3Au(s) + Cr3 +(aq)E∘cell=E∘red-E∘oxi=1.69-(-0.74)=2.43V

(b) Combine3×(2)and reversed of2×(3) :

3N2O(g)+6H++6e-+2Cr(s)→3N2(g)+3H2O(l)+2Cr3+(aq)+6e-

Cancel common terms to obtain the spontaneous reaction:

3N2O(g) + 6H++ 2Cr(s)→3N2(g) + 3H2O(l) + 2Cr3 +(aq) Ecella= Ereda- Eoxia= 1.77 - ( - 0.74) = 2.51V

(c) Combine (2) and reversed of 2×(1):

N2O(g)+2H++2e-+2Au(s)→N2(g)+H2O(l)+2Au+(aq)+2e-

Cancel common terms to obtain the spontaneous reaction:

N2O(g) + 2H++ 2Au(s)→N2(g) +H2O(l) + 2Au+(aq)EacellEaredEaoxi=1.77-(-0.74)=2.52V

Oxidising agents :N2O(g)>Au+(aq)>Cr3+(aq)

Reducing agents :Cr(s)>Au(s)>N2(g)

Therefore, the work done is

(a)3Au+(aq)+Cr(s)→3Au(s)+Cr3+(aq);E∘cell=2.43V

(b)3N2O(g)+6H++2Cr(s)→3N2(g)+3H2O(l)+2Cr3+(aq);E∘=2.51V

(c)N2O(g)+2H++2Au(s)→N2(g)+H2O(I)+2Au+(aq);E∘cell=0.08V

Oxidising agents:N2O(g)>Au+(aq)>Cr3+(aq)

Reducing agents :Cr(s)>Au(s)>N2(g)

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