Question

 L solution and a sodium sulfate solution use as a salt bridge solution. The silver electrode is positive relative to the chromium electrode. Write the cell half- reactions, overall reaction and cell notation. Calculate the following:

A) Gibbs Free – Energy               B) Electrode potential of the cell.

                     (Note: when calculating the molarity do not include the spectator ions)

Answer 1

A galvanic cell consists of chromium, Cr bar in a chromic sulfate solution, Cr2(SO4)3 containing 13 g and a silver, Ag bar in a silver sulfate solution (Ag2SO4 ) containing 37.8 g in a 3 L solution, and a sodium sulfate solution used as a salt bridge solution. The silver electrode is positive relative to the chromium electrode. Write the cell half-reactions, overall reaction, and cell notation. Calculate the following:

A) Gibbs Free – Energy

B) Electrode potential of the cell.

Molarity = No. of moles / Volume of solution (L).

No. of moles = Given mass / Molar mass

Molarity = Given mass / Molar mass x Volume of solution (L)

Molarity of Cr₂(SO₄)₃ $= \frac{13 \;g}{392.16 \;g/mol \times 3\; L} = 0.011 \;M$

Molarity of Ag₂SO₄ $= \frac{37.8 g }{ 311.799 \; g/mol \times 3\; L} = 0.04 \;M$

$E^0_{Cr3+ /Cr} = - 0.74 \; V \\ E^0_{Ag +/Ag} = 0.80 \;V$

The standard electrode potential of Ag is higher than Cr, therefore, Ag is the cathode and Cr is the anode.

Cell half reactions:

Anode reaction: Cr (s) → Cr ³⁺ (aq) + 3e^-

Cathode reaction: 3Ag⁺ (aq) + 3e- → 3Ag(s).

Overall reation: Cr(s) + 3Ag⁺(aq) → 3Ag(s) + Cr³⁺(aq)

Cell Notation:

Cr(s) / Cr(0.011 M) // Ag + (0.04 M) / Ag (s)

A. Gibbs Free energy

$\Delta G = - nFE_{Cel}l$

n is the number of electrons transferred

F is faraday's constant

$E_{Cell}$ is the cell potential

$E_{Cell} = E^0_{Cell} - (0.0591 \;V/ n) log([Anode]/[Cathode]) \\ E_{Cell}= E^0_{Cathode} - E^0_{Anode} \\= E^0_{Ag+/Ag} - E^0_{Cr 3+ /Cr} \\ = 0.80\; V - (- 0.74 V) \\= 1.54 \;V \\ E_{Cell} = 1.54 \;V - (0.0591 \;V / 3) log ([Cr 3+ ] / [Ag + ]^3 ) \\ E_{Cell} = 1.54 \;V - (0.0591 \;V / 3) log [0.011 / (0.04)^3 ] \\ E_{Cell} = 1.54 \;V - (0.0591 \;V / 3) log(171.875) \\ E_{Cell} = 1.54 \;V - 0.044 V = 1.496 V \\ \Delta G = - nFE_{Cell} \\ \Delta G = - 3 \times 96485 \times 1.496 = 433024.68 \;J/mol = 433.025 \; kJ/mol$

B. Electrode potential of the cell, $E_{Cell} = 1.496 \; V$