Question

Answer the following questions related to the given electrochemical cell.

IO⁻(aq) + H₂O(l) + 2e⁻ ⇌ I⁻(aq) + 2OH⁻(aq)
E° = 0.485 V

2NO(g) + H₂O(l) + 2e⁻ ⇌ N₂O(g) + 2OH⁻(aq)
E° = 0.760 V

3. The other cell compartment is comprised of a Pt electrode in a solution containing NO(g) at a pressure of 0.543 atm and N₂O(g) at a pressure of 0.345 atm at a temperature of 353.7 K. The concentration of OH⁻ is 8.67 × 10^-1 M.

(a) Choose the appropriate complete Nernst equation below for this half cell.

Remember that molarity and pressure are relative to 1 M and 1 atm, and that solids and liquids have a ratio of 1.

E = E° +	RT	ln	PNOa(1)b
	nF		PN2Oc[OH−]d

E = E° +	RT	ln	PNOa
	nF		PN2Oc

E = E° +	RT	ln	PN2Oc
	nF		PNOa

E = E° +	RT	ln	PN2Oc[OH−]d
	nF		PNOa(1)b

Tries 0/3

(b) What is E_cell (in V) for the NO/N₂O half cell? Report your answer to three decimal places in standard notation (i.e., 0.123 V).

Tries 0/3

4. Under the conditions described in questions 2 and 3:

(a) The half cell containing I⁻/IO⁻ is the  cathode anode

(b) The half cell containing NO/N₂O is the  cathode anode

(c) What is E_cell (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V).

Answer 1

3(a) for half cell the Ecell will be

Ecell = E⁰cell - RT/ nF[ ln Q]

Q = [OH-]²pN₂O /p²NO

Or Ecell = E⁰cell - RT/ nF[ln ( [OH-]²pN₂O /p²NO)]

Ecell = E⁰cell + RT/ nF[ln ( p²NO/ [OH-]²pN₂O)]

so last option is correct

b) Ecell = E0cell + RT/nF[ln ( p²NO/ [OH-]²pN₂O)]

Ecell = 0.760 + 0.0592 /2 log (p²NO/ [OH-]²pN₂O)

Ecell = 0.760 + 0.0296 log [(0.543)² / 0.345 X (0.867)²]

Ecell = 0.760 + 0.0296 X 0.0557

Ecell = 0.762 V

4) a) anode is I⁻/IO⁻

b) cathode is NO/N₂O

c) E⁰cell = E⁰cathode - E⁰anode

E⁰cell = 0.760 - 0.485 = 0.275 V