Answer the following questions related to the given electrochemical cell. 1. Ans
ID: 1006815 • Letter: A
Question
Answer the following questions related to the given electrochemical cell.
1. Answer the following questions under standard conditions.
(a) The half cell containing Cr/Cr2+ is the anode
(b) The half cell containing H+/H2 is the cathode
(c) What is E°cell (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V). 0.913 V
2. One cell compartment is comprised of an Cr electrode in a 1.31 × 10-5 M solution of Cr2+ at T = 348.2 K.
(a) Complete the Nernst equation below for this half cell by filling in the values and units where appropriate.
Important Note: Enter the values into the equation by following the order of the half reaction shown above.
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 Cr/Cr2+ half cell? Report your answer to three decimal places in standard notation (i.e., 0.123 V).
3. The other cell compartment is comprised of a Pt electrode in a 2.29 M solution of H+ and H2 pressure of 0.788 atm at T = 348.2 K.
(a) Complete the Nernst equation below for this half cell by filling in the values and units where appropriate.
Important Note: Enter the values into the equation by following the order of the half reaction shown above.
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 H+/H2 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 Cr/Cr2+ is the (cathode OR anode )
(b) The half cell containing H+/H2 is the (cathode OR anode )
(c) What is Ecell (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V).
Cr2+(aq) + 2e Cr(s)E° = -0.913 V 2H+(aq) + 2e H2(g)
E° = 0.000 V
Explanation / Answer
1. (a) The half cell containing Cr/Cr2+ is the anode
(b) The half cell containing H+/H2 is the cathode
(c) What is E°cell (in V)? Report your answer to three decimal places in standard notation
E0cell = 0.913 V
2. a) Ecell = E0cell - RT/nF ln(Cr^2+/1)
= -0.913-(8.314*348.2/(2*96500))ln(1.31*10^-5)
= -0.744 V
3. Ecell = E0cell - RT/nF ln(pH2/(H+)^2)
= 0-(8.314*348.2/(2*96500))ln(0.788/2.29^2)
= 0.028 V
4. a. anode
b. cathode
c. Ecell = 0.028--0.744 = 0.772 V
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