Answer the following questions related to the given electrochemical cell lr3-(aq

Question

Answer the following questions related to the given electrochemical cell lr3-(aq) + 3e-= Ir(s) E' = 1.156 V 2H (aq)+2e-H2lg) E. 0.000 V 1. What is E cell (in V)? Report your answer to two three decimal places in standard notation (i.e, 0.123 V) 1.156 V You are correct. Your receipt no. is 162-7561Previous Tries 2. The electrochemical cell is comprised of a Ir electrode in a 2.80 x 104 M solution of Ir3 (aq) coupled to a Pt electrode in a solution containing H (aq) where the pH of the solution is 2.10 and the partial pressure of H2(g) s 0.696 atm. The temperature of the cell is held constant at 25°C (a)What is Ecell (in V) for the electrochemical cell? Report your answer to three decimal places in standard notation (i.e, 0.123 V) Submit Answer Tries 1/3 Previous Tries (b) If the pH of the the cell compartment on the right is decreased to 0.96 while the other partial pressures and concentrations remain the same as in question 2, what will be the new Ecell (in V)? Report your answer to three decimal places in standard notation (i.e 0.123 V) Submit AnswerTries 0/3

Explanation / Answer

1. E0cell = E0cathode - E0anode

          = 1.156-0

          = 1.156 v

2. cell reaction : 2Ir3+(s) + 6H+(aq) <----> 2Ir(aq) + 3H2(g)

   Ecell = E0cell - (0.0591/n)log pH2^3/[H+]^6*[Ir3+]^2

         = 1.156 - (0.0591/6)log(0.696^3/((10^-2.1)^6*0.00028^2))

         = 0.966 V

         = 0.406 v

3. if pH = 0.96

Ecell = E0cell - (0.0591/n)log pH2^3/[H+]^6*[Ir3+]^2

         = 1.156 - (0.0591/6)log(0.696^3/((10^-0.96)^6*0.00028^2))

         = 1.034 v

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