*The Nernst Equation and pH*

Sulfuric acid is a very strong acid that can act as an oxidizing agent at high concentrations (low pH). Under standard conditions, sulfuric acid has a low reduction potential,

SO4^2-(aq) + 4H^+(aq) + 2e^- <=> SO2(g) + 2H2O(l),.....+0.20V

which means it cannot oxidize any of the halides F2, Cl2, Br2, or I2. If the H^+ ion concentration is increased, however, the driving force for the sulfuric acid reduction is also increased according to Le Châtelier's principle. Sulfuric acid cannot oxidize the fluoride or chloride anions, but it can oxidize bromide and iodide anions when there are enough H^+ ions present. The standard reduction potentials of the halogens are as follows:

F2 + 2e^- --> 2F^-,...+2.87V
Cl2 + 2e^- --> 2Cl^-,...+1.36V
Br2 + 2e^- --> 2Br^-,...+1.07V
I2 + 2e^- --> 2I^-,...+0.54V

The Nernst equation allows us to determine what nonstandard conditions allow the reaction to occur (have a positive E value).

THE ACTUAL QUESTION:
At 71.0^C, what is the maximum value of Q needed to produce a non-negative E value for the reaction:

SO4^2-(aq) + 4H^+(aq) + 2Br^-(aq) <=> Br2(aq) + SO2(g) + 2H2O(l)

In other words, what is Q when E=0 at this temperature?

Express your answer numerically using two significant figures.

Q = ___________?

3.2*10^-26

To determine the maximum value of Q needed to produce a non-negative E value for the given reaction at 71.0°C, we need to use the Nernst equation. The Nernst equation relates the standard reduction potential with the reaction quotient (Q) and the temperature.

The Nernst equation is given by:

E = E° - (RT / nF) * ln(Q)

Where:
E = cell potential under nonstandard conditions
E° = standard reduction potential
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
n = number of moles of electrons transferred in the balanced equation
F = Faraday's constant (96485 C/mol)
ln = natural logarithm
Q = reaction quotient

In this case, we want to find Q when E = 0 at 71.0°C.

First, let's find the standard reduction potential (E°) for the given reaction by subtracting the standard reduction potential of the reactants from the standard reduction potential of the products:

E° = (2 Br^-) - (1 SO4^2-) - (4 H^+) = (2 * (+1.07V)) - (+0.20V) - (4 * 0)

E° = +2.14V - 0.20V = +1.94V

Now, we can rearrange the Nernst equation to solve for Q:

E = E° - (RT / nF) * ln(Q)

0 = +1.94V - (RT / (2 * 96485 C/mol)) * ln(Q)

Since we are solving for Q when E = 0, we can simplify the equation to:

ln(Q) = (1.94V * 2 * 96485 C/mol) / (R * T)

Note: It's important to convert the units of temperature to Kelvin (°C to K) before plugging it into the equation.

ln(Q) = (1.94V * 2 * 96485 C/mol) / (8.314 J/(mol·K) * (71.0 + 273.15) K)

Now, we can solve for ln(Q) by dividing both sides of the equation by the constant on the right side:

ln(Q) = (1.94V * 2 * 96485 C/mol) / (8.314 J/(mol·K) * (71.0 + 273.15) K)

Calculating this value using a calculator, we get:

ln(Q) ≈ 17.479

To find Q, we need to take the exponential of both sides:

Q ≈ e^(17.479)

Calculating this using a calculator, we get:

Q ≈ 22750084518.91

Therefore, the maximum value of Q needed to produce a non-negative E value for the given reaction at 71.0°C is approximately 2.28 x 10^10 (rounded to two significant figures).

Q ≈ 2.28 x 10^10