Answer:
[tex]\boxed{5.5 \times 10^{-28}}[/tex]
Explanation:
We must use the Nernst equation
[tex]E = E^{\circ} - \frac{RT}{zF}\ln Q[/tex]
Step 1. Calculate E°
SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ ⇌ SO₂(g) + 2H₂O(ℓ)
2Br⁻⇌ Br₂(aq) + 2e⁻
SO₄²⁻(aq) + 4H⁺(aq) + 2Br⁻(aq) ⇌ Br₂(aq) + SO₂(g) + 2H₂O(ℓ)
E° = 0.17 - 1.0873 = -0.92 V
Step 2. Calculate Q
E = 0 V
E° = -0.92 V
R = 8.314 J·K⁻¹mol⁻¹
T = 66 °C
n = 2
F = 96 485 C/mol
Calculations:
T = 66.0 + 273.15 = 339.15 K
[tex]0 = -0.92 - \dfrac{8.314 \times 339.15}{2 \times 96 485}\ln Q\\\\0.92 = -0.0146\times \ln Q \\\\\ln Q = -\dfrac{0.92}{0.0146} = -62.8\\\\Q = e^{-62.8} =\boxed{5.5 \times 10^{-28}}\\[/tex]
The value of q when e = 0 at the given temperature in the question is :
Q = 1.3 * 10⁻²⁶
Given that
E = 0, ΔG = -nFE,
therefore ΔG = 0
Also
Given that
ΔG = ΔG° + RTIn q
ΔG° = - RTIn q
Hence ; Q = e^ (nFE°cell / RT) -- ( 1 )
where : n = 2, F = 96500, E°cell = -0.87 volt, R = 8.314, T = 339 k
insert values into equation ( 1 )
Q = 1.3 * 10⁻²⁶
Note : E°cell = reduction half reaction + oxidation half reaction
= 0.20 volt - 1.07 volt = -0.87 volt.
Hence we can conclude that The value of q when e = 0 at the given temperature in the question is : Q = 1.3 * 10⁻²⁶
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