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  4. A fuel cell involves combustion of the butane at at 1 atm and 298 K mathrmC4 mathrmH10( mathrm~g)+(13/2) mathrmO2( mathrm~g) arrow 4 mathrmCO2( mathrm~g)+5 mathrmH2 mathrmO(ℓ) Δ G° =-2744 kJ/mole The value of Ec e l lo Report your answer by rounding it upto nearest whole number.

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Q. A fuel cell involves combustion of the butane at at 1 atm and 298 K

$\mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(\ell)$

$\Delta G^\circ =-2744$ kJ/mole

The value of $E_{c e l l}^{o}$

Report your answer by rounding it upto nearest whole number.

NTA AbhyasNTA Abhyas 2020Electrochemistry

Solution:

$C_{4}H_{10}\left(g\right)+\frac{13}{2}O_{2}\left(g\right) \rightarrow 4\left(CO\right)_{2}\left(g\right)+5H_{2}O\left(l\right)$

Here total change in oxidation number of C-atoms $=+16-\left(\right.-10\left.\right)=+26$

It means total number of $e^{-}$ involved = 26

As $E_{c e l l}^{o}=\frac{- \Delta G ^\circ }{n F}=\frac{- \left(\right. - 2744 \left.\right) \times 1000}{26 \times 96500}$

$=+1.09V$

$\approx 1.0V$