A fuel cell involves combustion of the butane at 1 atm and 298 K C4H10g - 132O2g - 4CO2g - 5H2Ol G- - -2746 kJ-molWhat is E- of a cell-

The correct option is C + 1.09 V In the reaction 10 + 10C_4H_10(g) + 132 O_2(g) →+4 44CO_2(g) + 5H_2O(l) Change in oxidation number of carbon + ...

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Standard XII

Chemistry

EMF

A fuel cell i...

Question

A fuel cell involves combustion of the butane at $1 a t m$ and $298 K$
$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l) △ G^{\circ} = - 2746 k J / m o l$
What is $E^{\circ}$ of a cell?

+ 4.74 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 C $+ 1.09 V$
In the reaction
$- 10 + 10 C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to + 4 - 4 4 C O_{2} (g) + 5 H_{2} O (l)$
Change in oxidation number of carbon $+ 16 - (- 10) = + 26$
$∴$ Number of electrons involved in cell process will be $26$ .
$E^{\circ} = \frac{- △ G^{\circ}}{n F} = - \frac{(- 2746) \times 1000}{26 \times 96500} = + 1.09 V$ .

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Similar questions

Q. 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:

Q. A fuel cell develops an electrical potential from the combustion of butane at

1 b a r

and

298 K

C_{4} H_{10} (g) + 6.5 O_{2} (g) ⟶ 4 C O_{2} (g) + 5 H_{2} O (l); △_{r} G^{\circ} = - 2746 k J / m o l

What is

E^{\circ}

of a cell?

Q. A fuel cell develops an electrical potential from the combustion of butane at 1 bar and

298 K . C_{4} H_{10} (g) + {6.50}_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (ℓ) Δ_{r} G^{\circ} = - 2746

KJ/mole The

E^{\circ}

of the cell is ?

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

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A fuel cell involves combustion of the butane at 1 atm and 298 KC4H10(g)+132O2(g)→4CO2(g)+5H2O(l)△G∘=−2746 kJ/molWhat is E∘ of a cell?

The correct option is C +1.09 VIn the reaction−10+10C4H10(g)+132O2(g)→+4−44CO2(g)+5H2O(l)Change in oxidation number of carbon +16−(−10)=+26∴ Number of electrons involved in cell process will be 26.E∘=−△G∘nF=−(−2746)×100026×96500=+1.09 V.

A fuel cell involves combustion of the butane at $1 a t m$ and $298 K$
$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l) △ G^{\circ} = - 2746 k J / m o l$
What is $E^{\circ}$ of a cell?