-For each of the following reactions, write a balanced equation, calculate the s

Question

-For each of the following reactions, write a balanced equation, calculate the standard emf, calculate ?G? at 298 K, and calculate the equilibrium constant K at 298 K.

-Part A

Aqueous iodide ion is oxidized to I2(s) by Hg22+(aq).

Express your answer as a chemical equation. Identify all of the phases in your answer.

-Part B

Express your answer using three significant figures

-Part C

In acidic solution, copper (I) ion is oxidized to copper (II) ion by nitrate ion.

Express your answer as a chemical equation. Identify all of the phases in your answer.

Express your answer with two significant figures.

-Part D

In basic solution, Cr(OH)3(s) is oxidized to CrO2?4(aq) by ClO?(aq).

Express your answer as a chemical equation. Identify all of the phases in your answer.

-Part E

Express your answer using 3 significant figures.

Explanation / Answer

   Balancing of Redox reaction by Ion - exchange Mehod :-

Step1:- write and balance the oxidation and reduction half cell reactions separetely .

Step2:- First balance the atoms other than hydrogen and oxygen .

Step3:- balance the oxygen atom by adding sufficient number of moles of H2O in the oxygen deficient side .

Step4:- balance the hydrogen atom by adding sufficient number of H+ ions in the hydrogeb deficient side .

Step5:- balance the total charge on both sides by adding sufficient number of electrons .

Step6:-Now add oxidation and reduction half cell reaction in such a way that resulting (overall) equation does not contain any electron.

Part A:-

Oxidation half cell reaction (at anode ) :-

2I- (aq) ------------------> I2 + 2e- ,E0oxidation = - 0.535 V       .............(1)

Reduction half cell reaction(at cathode ) :-

Hg22+(aq) +    2e- ----------------> 2 Hg(l) , E0reduction = +0.796 V    ..................(2)

add equation (1) and (2), we have overall balanced equation:

2 I-(aq)   +   Hg22+(aq) ----------> I2(s)   +   2 Hg (l)  

E0cell= E0oxidation + E0reduction = -0.535V + 0.796V = 0.261V

E0cell = 0.261 V

also , deltaG0 = -nFE0cell

deltaG0= - 2 x 96500 C x 0.261 V

deltaG0 = - 50373 J/mol

deltaG0 = - 50.373 KJ/mol

because , deltaG0 = - 2.303 RT log K

- 50373 J/mol = - 2.303 x 8.314 J/Kmol x 298 K log K

log K = 50373 / 2.303 x 8.314 x 298

log K = 8.828

K = 108.828

K = 6.73 x 108

Part B:-

Eocell = 0.261 V

deltaG0 = - 50.4 KJ /mol

K = 6.73 x 108

Part C:-

Oxidation half cell reaction (at anode ) :-

Cu+ (aq) ------------------> Cu2+(aq) + 1e- ,E0oxidation = -0.159 V       .............(1)

Reduction half cell reaction(at cathode ) :-

NO3-(aq) + 2H+(aq) + 1e- ----------------> NO2(g) + H2O(l) , E0reduction = +0.80 V    ..................(2)

add equation (1) and (2), we have overall balanced equation:

Cu+(aq) + NO3-(aq) + 2 H+(aq) ---------> Cu2+(aq) + NO2(g) +   H2O

E0cell= E0oxidation + E0reduction = -0.159 V + 0.80 V = 0.64 V

E0cell = 0.64 V

also , deltaG0 = -nFE0cell

deltaG0= - 1 x 96500 C x 0.64 V

deltaG0 = - 61760 J/mol

deltaG0 = - 61.760 KJ/mol

because , deltaG0 = - 2.303 RT log K

- 61760 J/mol = - 2.303 x 8.314 J/Kmol x 298 K log K

log K = 61760 / 2.303 x 8.314 x 298

log K = 10.824

K = 1010.824

K = 6.67 x 1010

Eocell = 0.64 V

deltaG0 = - 6.2 x 101 KJ /mol

K = 6.8 x 1010

Part D:-

Oxidation half cell reaction (at anode ) :-

Cr(OH)3 + 5 OH- ------------------>CrO22- + 4H2O + 3e- ,E0oxidation = +0.13 V V       .............(1)

Reduction half cell reaction(at cathode ) :-

ClO- + H2O +    2e- ----------------> Cl- + 2 OH- , E0reduction = +0.89 V    ..................(2)

Multiply equation (1) by 2 and equation (2) by 3 and after that add equation (1) and (2), we have overall balanced equation:

2Cr(OH)3 + 3 ClO- + 4OH- ---------->2CrO42- + 3Cl- + 5H2O

E0cell= E0oxidation + E0reduction = +0.13 V + 0.89 V = 1.02 V

E0cell = 1.02 V

also , deltaG0 = -nFE0cell

deltaG0= - 6 x 96500 C x 1.02 V

deltaG0 = - 590580 J/mol

deltaG0 = - 590.58 KJ/mol

because , deltaG0 = - 2.303 RT log K

- 590580J/mol = - 2.303 x 8.314 J/Kmol x 298 K log K

log K = 590580 / 2.303 x 8.314 x 298

log K = 103.5

K = 10103.5

K = 3.16 x 10103

Part E:-

Eocell = 1.02 V

deltaG0 = - 5.91 x 102 KJ /mol

K = 3.16 x 10103

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.