Answer the following questions related to the given electrochemical cell. IO (aq

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.

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

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 Ecell (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V).

Explanation / Answer

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

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