A fuel cell involves combustion of the butane at 1 atm and 298 kelvin- C 4 H 10 g -13-2 O 2 g - 4 CO 2 g -5 H 2 O l - G -2746 kJ - molWhat is the E - of this cell-A- 0-547 VB- -1-09 VC- -4-37 VD- -4-76 V

The correct option is B +1.09 V 10C_4H_10(g) + 132O_2(g) → 4+16CO_2(g) + 5H_2O(l) Number of electrons involved = 16 ( 10)=26 Δ G^∘ = nFE_cell^∘ 2746 × 10^3 ...

Byju's Answer

Standard XII

Chemistry

Gibb's Energy and Nernst Equation

A fuel cell i...

Question

A fuel cell involves combustion of the butane at 1 atm and 298 kelvin.
$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$

$Δ G^{\circ} = - 2746 kJ/mol$
What is the $E^{\circ}$ of this cell:

+4.76 V

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0.547 V

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+1.09 V

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+4.37 V

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Solution

The correct option is B +1.09 V
${- 10 C}_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 + 16 C O_{2} (g) + 5 H_{2} O (l)$

Number of electrons involved $= 16 - (- 10) = 26$
$Δ G^{\circ} = - n F E_{cell}^{\circ}$
$- 2746 \times 10^{3} = - 26 \times 96500 \times E_{cell}^{\circ}$
$E_{cell}^{\circ} = 1.09 V$

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During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is
(a) $2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{2} O (l); Δ_{c} H = - 2658.0 k J m o l^{- 1}$
(b) $C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l); Δ_{c} H = - 1329.0 k J m o l^{- 1}$
(c) $C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l); Δ_{c} H = - 2658.0 k J m o l^{- 1}$
(d) $C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l); Δ_{c} H = + 2658.0 k J m o l^{- 1}$

C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) ⟶ 4 C O_{2} (g) + 5 H_{2} O (l); Δ H = - 2878 k J

Δ H

is the heat of combustion of butane gas.If true enter 1 else 0

A fuel cell involves combustion of the butane at 1 atm and 298 kelvin. C4H10(g)+132O2(g)→4CO2(g)+5H2O(l) ΔG∘=−2746 kJ/mol What is the E∘ of this cell:

The correct option is B +1.09 V−10C4H10(g)+132O2(g)→4+16CO2(g)+5H2O(l) Number of electrons involved =16−(−10)=26 ΔG∘=−nFE∘cell −2746×103=−26×96500×E∘cell E∘cell=1.09 V

A fuel cell involves combustion of the butane at 1 atm and 298 kelvin.
$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$

$Δ G^{\circ} = - 2746 kJ/mol$
What is the $E^{\circ}$ of this cell: