Answer the following questions related to following electrochemical cell. Au(OH)

Question

Answer the following questions related to following electrochemical cell.

Au(OH)3(aq) + 3H+(aq) + 3e? ? Au(s) + 3H2O(l)
E° = 1.450 V

2IO3?(aq) + 12H+(aq) + 10e? ? I2(s) + 6H2O(l)
E° = 1.195 V

1. Answer the following questions under standard conditions.

(a) What is E°cell (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V). 0.255 V

(b) Which one of the following chemical reactions describes the process occurring in the electrochemical cell?
Incorrect: 20Au(s) + 12IO3?(aq) + 36H2O(l) + 12H+(aq) ? 30Au(OH)3(aq) + 6I2(s)
Correct: 10Au(OH)3(aq) + 3I2(s) ? 10Au(s) + 6IO3?(aq) + 12H2O(l) + 6H+(aq)
Incorrect: 30Au(OH)3(aq) + 6I2(s) ? 20Au(s) + 12IO3?(aq) + 36H2O(l) + 12H+(aq)
Incorrect: 10Au(s) + 6IO3?(aq) + 12H2O(l) + 6H+(aq) ? 10Au(OH)3(aq) + 3I2(s)

(c) What is ?G° (in kJ/mol)? Report your answer to three significant figures in scientific notation (i.e., 1.23e2 kJ/mol). -7.38×102 kJ/mol

2. Consider the electrochemical cell where one of the half cells is comprised of a Pt electrode in a solution containing Au(OH)3(aq) at a concentration of 3.39 × 10-4 M and Au(s) with a H+ concentration of 2.85 × 10-5 M. The other half cell is comprised of a Pt electrode in a solution containing IO3?(aq) at a concentration of 6.19 × 10-1 M and I2(s) with a H+ concentration of 6.00 M. The temperature of the cell is 291.3 K.

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

(b) Which one of the following chemical reactions describes the process occurring in the electrochemical cell?
30Au(OH)3(aq) + 6I2(s) ? 20Au(s) + 12IO3?(aq) + 36H2O(l) + 12H+(aq)
20Au(s) + 12IO3?(aq) + 36H2O(l) + 12H+(aq) ? 30Au(OH)3(aq) + 6I2(s)
10Au(OH)3(aq) + 3I2(s) ? 10Au(s) + 6IO3?(aq) + 12H2O(l) + 6H+(aq)
10Au(s) + 6IO3?(aq) + 12H2O(l) + 6H+(aq) ? 10Au(OH)3(aq) + 3I2(s)

(c) What is ?G (in kJ/mol)? Report your answer to three significant figures in scientific notation (i.e., 1.23e2 kJ/mol)

Explanation / Answer

1: (a), (b) and (c) are correct.

2: (a): Oxidation-half reaction: 3I2(s) + 18H2O(l) ---------> 6IO3(aq) + 36H+(aq) + 30e : E0(oxi) = - 1.195 V

Reduction-half reaction: 10Au(OH)3(aq) + 30H+(aq) + 10e ----------> 10Au(s) + 30H2O(l): E0(red) = 1.450 V

E(oxi) =  E0(oxi) - (0.0591/30)xlog [H+(aq)]36x[IO3-(aq)]6

=> E(oxi) = - 1.195 V - (0.0591/30)xlog(6.00)36x(0.619)6 = - 1.248 V

E(red) =  E0(red) - (0.0591/30)xlog(1 / [H+(aq)]30x[Al(OH)3(aq)]10

=> E(red) = 1.450 V - (0.0591/30)xlog(1 / (2.85x10-5)30x(3.39x10-4)10 = V

=> E(red) = 1.113 V

Hence E(cell) = E(oxi) + E(red) = -1.248 + 1.113 = - 0.135 V (answer)

(b): Since E(cell) is negative the reverse reaction will be feasible. Hence the cell reaction that occurs inside the cell is

10Au(s) + 6IO3(aq) + 12H2O(l) + 6H+(aq) 10Au(OH)3(aq) + 3I2(s)

(c): deltaG = - nFE(cell) = - 30 x 96500 x (-0.135) = +3.91x102 kJ/mol (answer)

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