Question

At 73.0 ∘C , what is the maximum value of the reaction quotient, Q, needed to produce a non-negative E value for the reaction

SO42−(aq)+4H+(aq)+2Br−(aq)⇌Br2(aq)+SO2(g)+2H2O(l)

In other words, what is Q when E=0 at this temperature? And the sovle for the pH?

Answer 1

As we know

E = E^o -( RT/nF )ln Q

Whre E = 0 (given)

R : gas constant = 8.314 J/mol.K

T is the temperature = 73^oC = 73 + 273 = 346 K

F = Faraday constant = 96500 J/mol.K

n = number of electrons involved = 2

and Two half reactions are:

2Br^- (aq)Br₂(aq) + 2e^- ( oxidized)

Reduction value for this is E = 1.07 V

and SO₄^2-(aq) + 4H⁺(aq) + 2e^- SO₂(g) + 2H₂O (l) (Reduced)

Reduction vlue for this half cell is E = 0.2

E^o = E_reduced - E_oxidized = 0.2 - 1.07 V = -0.87 V

0 = -0.87 V - (8.314 J/mol.K * 346 K) ln Q / ( 2* 96500 J/mol.K)

Q = 4.57 * 10^-26

For pH :