I need help with these electrochemical cell questions please Answer the followin
ID: 948768 • Letter: I
Question
I need help with these electrochemical cell questions please
Answer the following questions related to the given electrochemical cell H2AlO3 (a)H20()3e-Al(s) +40H (aq) Eo -2.330 V 2e-N20 (g) 20H (aq) 2NO(g) H20(I) Eo = 0.760 V 1. Answer the following questions under standard conditions (a) The half cell containing Al/H2AlO3 is the anode (b) The half cell containing NO/N20 is the cathode (c) What is Eocell (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V). 3.090 V You are correct Previous Tries Your receipt no. is 160-956 2. One cell compartment is comprised of a Pt electrode in a solution containing H2AIO3-(aq) at a concentration of 4.54 × 10-4 M and Al(s) at a temperature of 294.9 K. The concentration of OH-is 6.00 × 10-5 M (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 RT [ox]a 8.314 Jmol-"K- In nF [red] 96490 Cmor1 Submit Answer Tries 0/3 (b) What is Ecell (in V) for the Al/H2AlO3 half cell? Report your answer to three decimal places in standard notation (i.e., 0.123 V) Submit Answer Tries 0/3 3. The other cell compartment is comprised of a Pt electrode in a solution containing NO(g) at a pressure of 0.498 atm and N2O g at a pressure of 0.512 atm at a temperature of 294.9 K. The concentration of OH-is 6.73 × 10 (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 1Explanation / Answer
2.
a. Nernst equation,
E = -2.330 + [(8.314 x 294.9/3 x 96485)ln(4.54 x 10^-4/(6 x 10^-5)^2)]
b. Ecell = -2.230 V
3.
a. Nernst equation,
Ecell = 0.760 + [(8.314 x 294.9/3 x 96485)ln((0.498)^2/(0.512)(6.73 x 10^-5)^2)]
b. Ecell = 0.916 V
4. According to the given data,
(a) The half cell Al/H2AlO3- is anode
(b) the half cell NO/N2O is cathode
(c) Ecell = 3.146 V