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swati

At 298K the free energy of formation of H2O(l) is -237.23 kJ.while that of its ionization to hydrogen and hydroxyl ions is 79.71kJ.what will be the emf at 298K of the cell Pt|H2(1atm) | H+ || OH- |O2(1ATM) |Pt. IF the heat of formation at 298K of H2O(l) is -285.85kJwhile its heat of ionization is56.9kJ.calculate the temperature coeffecient at 298K for the above cell.

smriti.mathur

Left side of the cell is oxidation half cell -

1/2 H_{2 (g)}

H⁺ + e^- .................(1)

Right side is reduction half cell -

1/2 O_{2 (g)} + H₂O + 2e^-

2OH^- ................(2)

Multiply eq.(1) by 2

H_{2 (g)}

2H⁺ + 2e^-

Cell Reaction -

H_{2 (g)} + 1/2 O_{2 (g)} + H₂O

2H⁺ + 2OH^-

G = - nEF

G = Free energy of cell

n = no. of electrons involved in the redox reaction

E = E.M.F of cell

F = Faraday

H_{2 (g)} + 1/2O_{2 (g)}

H₂O ;

G = -237.23KJ/mol .............(3)

H₂O

H⁺ + OH^- ;

G = +79.71KJ/Mol .................(4)

Multiply eq(4) by 2 and add with eq (3)

H_{2 (g)} + 1/2 O_{2 (g)} + H₂ O

2H⁺ +2OH^- ;

G = -77.81KJ/mol(multiply by 1000 to convert in joules)

So, -77.81 = -2 * E * 96500

E = 0.4 volts

H_{2 (g)} + 1/2O_{2 (g)}

H₂O ;

H = -285.85KJ/mole .......(5)

H₂O

H⁺ + OH^- ;

H = 56.9KJ/mol .................(6)

Multiply eq(6) by 2 and add with eq(5)

H_{2 ( g)} + 1/2O_{2 (g)} + H₂O

2H⁺ + 2OH^- ;

H = -172.0KJ/mol

Now,

Temperature Coefficient = dE/dT

E = -

H/nF + T(dE/dT)

T =298K

Putting the values of E,

H, n and F, the temp. coefficient can be calculated. (Convert

H into joules by multiplying with 1000).

swati

well! i can only say that u replied very soon.

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