A solid piece of graphite was used as the electrode in each side of the cell (GR

Question

A solid piece of graphite was used as the electrode in each side of the cell (GRAPHITE IS AN INERT ELECTRODE AND DOES NOT UNDERGO OXIDATION OR REDUCTION)

One side of the cell had a solutionof 0.1 M KI/0.001 M I2 (aq)

The other side of the cell had a solutio of 0.01M KI/0.001 M I2 (aq)

PLEASESHOW WORK FOR PART G IN ORDER TO GET POINTS.... THANK YOU

a. Identify the anode electrode.

b. Identify the cathode electrode.

c. Write the balanced oxidation half reaction and its standard oxidation potential.

d. Write the balanced reduction half reaction and its standard oxidation potential.

e. Write the overall balanced redox reaction and its standard cell potential.

f. Write the line notation for the cell.

g. Calculate the theoretical cell voltage using the standard cell potential and the solution concentrations.

Explanation / Answer

(a) Anode: C(s) | I-(aq, 0.1 M), I2(aq, 0.001M)

(b) Cathode: I2(aq, 0.001 M), I-(aq, 0.01 M) | C(s)

(c) Oxidation half reaction:

2 I-(aq, 0.1 M) => I2(aq, 0.001 M) + 2 e-, Eo(ox) = -0.54 V

(d) Reduction half reaction:

I2(aq, 0.001 M) + 2 e- => 2 I-(aq, 0.01 M), Eo(red) = +0.54 V

(e) OVerall reaction:

2 I-(aq, 0.1 M) => 2 I-(aq, 0.01 M)

Eo(cell) = Eo(ox) + Eo(red) = -0.54 + 0.54 = 0.00 V

(f) Cell notation:

C(s) | I-(aq, 0.1 M), I2(aq, 0.001M) || I2(aq, 0.001 M), I-(aq, 0.01 M) | C(s)

(g) Moles of electrons transferred n = 2

Faraday constant F = 96485 C/mol

Molar gas constant R = 8.314 J/mol.K

Temperature T = 25 deg C = 298.15 K

Nernst equation:

E(cell) = Eo(cell) - RT/nF ln([I-(0.01 M)]^2/[I-(0.1 M)]^2)

= 0.00 - 8.314 x 298.15/(2 x 96485) x ln(0.01^2/0.1^2)

= 0.0592 V (or approximately 0.059 V)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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