Question

For an element X, the following standard reduction potentials exist at a temperature of our study:

X^3+(aq) + 2 e– → X^+(aq) ε1° = 1.192 V

X^4+(aq) + 3 e– → X^+(aq) ε2° = 1.241 V

If [X^+] = 0.170 M, [X^3+] = 5.670 M, and [X^4+] = 0.174 M, at what temperature (in °C) will the galvanic/voltaic cell made from the above two half-reactions begins to become nonspontaneous?

Answer 1

The given half cell reactions are-

X³⁺(aq) + 2 e– → X⁺(aq) ε1° = 1.192 V

X⁴⁺(aq) + 3 e– → X⁺(aq) ε2° = 1.241 V

Here both the reactions are writeen in Reduction form. But we know in an electrochemical cell, both Reduction (i.e gain if electron) and Oxidation (i.e loss of electron) takes place

So according to the rule, the half cell which has the larger ε° value is the actual reduction half cell where as the other is the Oxidation one. So here our

Reduction half cell:   X³⁺(aq) + 2 e– → X⁺(aq) ε1° = 1.192 V

Oxidation half cell:   X⁴⁺(aq) + 3 e– → X⁺(aq) ε2° = 1.241 V

Or

Reduction half cell:   X³⁺(aq) + 2 e– → X⁺(aq) ε1° = 1.192 V

Oxidation half cell:   X⁺(aq) → X⁴⁺(aq) + 3 e– ε2° = 1.241 V

Now to form the overall reaction, we have to balance the reactions by equating the electrons. i.e

Reduction half cell:   X³⁺(aq) + 2 e– → X⁺(aq) ]*3 ε1° = 1.192 V * 3

Oxidation half cell:   X⁺(aq) → X⁴⁺(aq) + 3 e– ]*2 ε2° = 1.241 V * 2

Or

Reduction half cell: 3X³⁺(aq) + 6 e– → 3X⁺(aq) ε1° = 3.576‬ V

Oxidation half cell: 2X⁺(aq) → 2X⁴⁺(aq) + 6 e– ε2° = 2.482 V

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Overall reaction:   3X³⁺(aq) → X⁺(aq) + 2X⁴⁺(aq)   ε° = ε°(Reduction) - ε°(Oxidation)

= 3.576‬ V - 2.482 V

= 1.094 V

Now for a galvanic cell to have a non-spontaneous reaction, the value of E_cell should always be negative.

Again we can calculate E_cell from Nernst equation as-

E_cell = ε° - RT/nF * ln [Q]

Where

R = gas constant = 8.314 J/mol.K

T = Given temperature in kelvin

n = moles of electron transfered = here it is 6

F = Faraday's constant = 96485 C/mol

Q = Reaction quotient

= [X⁺] * [X⁴⁺]² / [X³⁺]³  

= [0.170] * [0.174]² / [5.670]³  

= 2.8 * 10^-5  

Now putting these values-

E_cell = ε° - RT/nF * ln [Q]

= 1.094 V - [(8.314 J/mol.K) * T/ (6 mols * 96485 C/mol)] * ln [2.8 * 10^-5]

= 1.094 V - [(8.314 J/mol.K) * T/ (5,78,910‬ C)] * (-10.48)

Now for E_cell to be negative, we can have-

[(8.314 J/mol.K) * T/ (5,78,910‬ C)] * (-10.48) >  1.094 V

   -(1.5 * 10^-5) V * T > 1.094 V

-T > 1.094 V / (1.5 * 10^-5) V

-T > 72,933

T > -72,933 K

So in ^oC,   T > -72,933 - 273

   T > -73,206 ^oC

i.e minimum temperature in ^oC should be  -73,206 ^oC

= 1.094 V + (1.5 * 10^-5) J/mol.C * T

= 1.094 V + (1.5 * 10^-5) V * T (1 J/mol.C = 1 V)

Now for E_cell to be negative, (1.5 * 10^-5) V * T should be a negative number higher value than 1.094 V . So that the overall E_cell will have a negative value

So the minimum value of (1.5 * 10^-5) V * T can be -(1.095 V )

So if we take   (1.5 * 10^-5) V * T can be -(1.095 V )

That means in the least value, we can take (1.5 * 10^-5) V * T to be -1.093 and not more than that.

So if we take (1.5 * 10^-5) V * T = -1.093 V

Then T = -1.093 V/ (1.5 * 10^-5) V

= 72,867 K

So T = 72,867 - 273