JEE Advance - Chemistry (2020 - Paper 1 Offline - No. 15)
Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is
$${H_2}(g) + {1 \over 2}{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l)$$
The work derived from the cell on the consumption of 1.0 $$ \times $$ 10$$-$$3 mole of H2(g) is used to compress 1.00 mole of a monoatomic ideal gas in a thermally insulated 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 :
$${O_2}(g) + 4{H^ + }(aq) + 4{e^ - }\buildrel {} \over \longrightarrow 2{H_2}O(l),$$
$${E^o} = 1.23V$$
$$2{H^ + }(aq) + 2{e^ - }\buildrel {} \over \longrightarrow {H_2}(g),$$
$${E^o} = 0.00\,V$$
Use, $$F = 96500\,C\,mo{l^{ - 1}}$$, $$R = 8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$$.
$${H_2}(g) + {1 \over 2}{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l)$$
The work derived from the cell on the consumption of 1.0 $$ \times $$ 10$$-$$3 mole of H2(g) is used to compress 1.00 mole of a monoatomic ideal gas in a thermally insulated 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 :
$${O_2}(g) + 4{H^ + }(aq) + 4{e^ - }\buildrel {} \over \longrightarrow 2{H_2}O(l),$$
$${E^o} = 1.23V$$
$$2{H^ + }(aq) + 2{e^ - }\buildrel {} \over \longrightarrow {H_2}(g),$$
$${E^o} = 0.00\,V$$
Use, $$F = 96500\,C\,mo{l^{ - 1}}$$, $$R = 8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$$.
Answer
13.32
Explanation
Vessel is insulated, thus q = 0
For the given reaction : $$\eqalign{ & {H_2}(g) + 1/2{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l); \cr & {E^o} = 1.23 - 0.00 = 1.23V \cr} $$
$$\Delta $$Go = $$-$$nFEo
= $$-$$2 $$ \times $$ 96500 $$ \times $$ 1.23 J/mol
Therefore, work derived from this fuel cell using 70% efficiency and on consumption of 1.0 $$ \times $$ 10$$-$$3 mol of H2(g)
= 2 $$ \times $$ 96500 $$ \times $$ 1.23 $$ \times $$ 0.7 $$ \times $$ 1 $$ \times $$ 10$$-$$3
= 166.17 J
This work done = change in internal energy (for monoatomic gas,
$${C_{V'm}} = 3R/2$$),
$$166.17 = n{C_V}{,_m}\Delta T$$
$$ \Rightarrow \Delta T = {{166.17 \times 2} \over {1 \times 3 \times 8.314}}$$
$$ \Rightarrow 13.32\,K$$
For the given reaction : $$\eqalign{ & {H_2}(g) + 1/2{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l); \cr & {E^o} = 1.23 - 0.00 = 1.23V \cr} $$
$$\Delta $$Go = $$-$$nFEo
= $$-$$2 $$ \times $$ 96500 $$ \times $$ 1.23 J/mol
Therefore, work derived from this fuel cell using 70% efficiency and on consumption of 1.0 $$ \times $$ 10$$-$$3 mol of H2(g)
= 2 $$ \times $$ 96500 $$ \times $$ 1.23 $$ \times $$ 0.7 $$ \times $$ 1 $$ \times $$ 10$$-$$3
= 166.17 J
This work done = change in internal energy (for monoatomic gas,
$${C_{V'm}} = 3R/2$$),
$$166.17 = n{C_V}{,_m}\Delta T$$
$$ \Rightarrow \Delta T = {{166.17 \times 2} \over {1 \times 3 \times 8.314}}$$
$$ \Rightarrow 13.32\,K$$
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