A concentration cell is constructed from two hydrogen electrodes connected by a

Question

A concentration cell is constructed from two hydrogen electrodes connected by a salt bridge. The pressure of H2 is 1.00 atm in each hydrogen electrode. One electrode is immersed in a solution with[H+] = 1.0 M and the other in a solution with 0.41 M NH3. What will be the value of Ecell? (Hint: First, determine the [H+] in the NH3 solution by considering the equilibrium problem.)

Enter your answer with 3 significant digits.
Enter scientific notation as 1.23E4.

NOTE: I previously thought the answer was 2.28*10^-2 which is incorrect. The feedback I got for this quesition was:

The expression for Kb is [NH4+] [OH]/[NH3] = 1.8 × 105. For a solution of NH3, one can set [NH4+] = [OH] =x to get x2/(C-x) = 1.8 × 105, and assuming x << C. This yields a simple expression to  solve for x = [OH]. [OH], can then be used to calculate [H+]. This value of [H+] can then be used, with other quantities, to calculate the cell potential using the Nernst equation.

PLEASE CALCULATE THE CORRECT ANSWER MANY THANKS

A concentration cell is constructed from two hydrogen electrodes connected by a salt bridge. The pressure of H2 is 1.00 atm in each hydrogen electrode. One electrode is immersed in a solution with[H+] = 1.0 M and the other in a solution with 0.41 M NH3. What will be the value of Ecell? (Hint: First, determine the [H+] in the NH3 solution by considering the equilibrium problem.)

Enter your answer with 3 significant digits.
Enter scientific notation as 1.23E4.

NOTE: I previously thought the answer was 2.28*10^-2 which is incorrect. The feedback I got for this quesition was:

Incorrect

The expression for Kb is [NH4+] [OH]/[NH3] = 1.8 × 105. For a solution of NH3, one can set [NH4+] = [OH] =x to get x2/(C-x) = 1.8 × 105, and assuming x << C. This yields a simple expression to  solve for x = [OH]. [OH], can then be used to calculate [H+]. This value of [H+] can then be used, with other quantities, to calculate the cell potential using the Nernst equation.

PLEASE CALCULATE THE CORRECT ANSWER MANY THANKS

Explanation / Answer

Kb = [NH4+][OH-]/[NH3]

let x be the change in concentration at equilibrium

1.8 x 10^-5 = x^2/0.41

x = [OH-] = 2.72 x 10^-3 M

[H+] = 1 x 10^-14/2.72 x 10^-3 = 3.68 x 10^-12

Using Nernst equation for concentration cell,

E = -0.0592 log(3.68 x 10^-12/1) = 0.677 V = 6.77 x 10^-1 V

for the cell

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