Consider a 70 % efficient hydrogen-oxygen fuel cell working under… - Vidyadip

MCQ

Consider a $70 \%$ efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and $298 K$. Its cell reaction is

$H _2( g )+\frac{1}{2} O _2( g ) \rightarrow H _2 O (\ell)$

The work derived from the cell on the consumption of $1.0 \times 10^{-3} mol$ of $H _2( g )$ is used to compress $1.00 mol$ of a monoatomic ideal gas in a thermally insulted container. What is the change in the temperature (in $K$ ) of the ideal gas ?

The standard reduction potentials for the two half-cells are given below.

$\left. O _2( g )+4 H ^{+} \text {(aq. }\right)+4 e ^{-} \rightarrow 2 H _2 O (\ell), E ^{\circ}=1.23 V,$

$\left.2 H ^{+} \text {(aq. }\right)+2 e ^{-} \rightarrow H _2( g ), E ^{\circ}=0.00 V.$

Use $F =96500 C mol ^{-1}, R =8.314 J mol ^{-1} K ^{-1}$

✓
$13.32$
B
$13.35$
C
$13.40$
D
$13.45$

✓

Answer

Correct option: A.

$13.32$

a
$E _{\text {cell }}^0=1.23-0.00=1.23 V$

$\Delta G _{\text {cell }}^0=- nF E _{\text {cell }}^0=-2 \times 96500 \times 1.23 J$

$\therefore \quad$ Work derived from this fuel cell

$=\frac{70}{100} \times\left(-\Delta G _{\text {cell }}^0\right) \times 10^{-3}= xJ$

Since insulated vessel, hence $q=0$

From equation, for monoatomic gas,

$w =\Delta U \quad \Rightarrow \quad x = nC _{ V _{ m }} \Delta T \left\{ C _{ V _{ m }}=\frac{3 R }{2}\right\}$

or,$\quad \frac{70}{100} \times(2 \times 96500 \times 1.23) \times 10^{-3}=1 \times \frac{3}{2} \times 8.314 \times \Delta T$

$\therefore \quad \Delta T =13.32$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 2. Electrochemistry→2. Electrochemistry — all question types→Chemistry STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

A sample of milk splits after $60 \;min$. at $300\; \mathrm{K}$ and after $40 \;min$. at $400 \;\mathrm{K}$ when the population of lactobacillus acidophilus in it doubles. The activation energy (in $\mathrm{kJ} / \mathrm{mol}$ ) for this process is closest to ............. $\mathrm{kJ/mole}$

(Given, $\mathrm{R}=8.3\; \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$$, \ln \left(\frac{2}{3}\right)=0.4$ $\left.e^{-3}=4.0\right)$

$\begin{gathered}
\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - {\text{ }}C - {\text{ }}O - H\xrightarrow{{NaH{{\mathop {CO}\limits^{14} }_3}}}\mathop {(A)}\limits_{(gas)} \xrightarrow[{(ii){H_3}{O^ \oplus }}]{{(i)PhMgBr}}(B)}
\end{array} \hfill \\
\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - S - O - H\xrightarrow{{NaH\mathop {C{O_3}}\limits^{14} }}\mathop {(C)}\limits_{(gas)} \xrightarrow[{(ii){H_3}{O^ \oplus }}]{{(i)PhMgBr}}(D)} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array} \hfill \\
\end{gathered}$

Product $(B)$ and $(D) $ in the above reaction are

An octahedral complex with no chelate ring of $M^{+3}$ with $NH_3$ and $NO_2^{\circleddash } $ ligands only have four $M-N$ linkage and two $M-O$ linkage, then such complex will show :-

A first order reaction has a rate constant of $2.303 \times 10^{-3} \;\mathrm{s}^{-1} .$ The time required for $40 \mathrm{g}$ of this reactant to reduce to $10\; \mathrm{g}$ will be.....$s$

[Given that $\left.\log _{10} 2=0.3010\right]$

Specific conductance of $0.2\, M$ electrolyte solution at $20\,^oC$ temperature is $2.48\times10^{-4}\, ohm^{-1}\, cm^{-1}$ then the molar conductance of solution will be ............ $\mathrm{ohm}^{-1} \mathrm{cm}^{2} \mathrm{mol}^{-1}$

The end formed between the interaction of two amino acid is:

The first ionisation potential of N, P, O and S are in the order of:

Standard electrode potentials of $Zn$ and $Fe$ are known to be $ (i) $ $ - 0.76\,V$ and $ (ii) $ $ - 0.44\,V$ respectively. How does it explain that galvanization prevents rusting of iron while zinc slowly dissolves away

What is the major product $"P"$ of the following reaction?

In which case, the order of acidic strength is not correct ?