Consider a $70 %$ efficient hydrogen-oxygen cell working under the standard condition at $1$ bar and $298$ K. Its cell reaction is
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
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 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
$O_{2} (g) + 4 H^{+} (aq) + 4 e^{-} \to 2 H_{2} O (l)$ , $E^{0}$ = $1.23 V$
$2 H^{+} (aq) + 2 e^{-} \to H_{2} (g)$ , $E^{0}$ = $0.00 V$
Use $F = 96500 {Cmol}^{- 1}$ , $R = 8.314 {Jmol}^{- 1} K^{- 1}$ .

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Solution

Step 1. Given data:
$F = 96500 {Cmol}^{- 1}$
$E^{0}$ $=$ $1.23 V$
$∆ U$ is change in internal energy
$C_{v, m}$ is the specific heat of monoatomic gas $(\frac{3}{2}) R$
Step 2. Calculating $∆ G^{0}$ :
$∆ G^{0} = - {nFE}^{0}_{cell}$
$n$ is the number of electrons
$n = 2$
$∆ G^{0}$ $=$ $- 2 \times 96500 \times 1.23 \times 1 \times 10^{- 3} \times 0.7$
$= - 166.17$ $J$
Step 3. Calculating $∆ T$ :
$∆ U = {nC}_{v, m} ∆ T$
$n$ is moles of gas
$n = 1$
$∆ T$ is change in temperature.
Since, the monoatomic ideal gas is placed in a thermally isolated container then, $∆ H = 0$
$w = ∆ U w = - ∆ G$
Hence,
$∆ U =$ $166.17 J$
Substituting given values
$166.17 = 1 \times \frac{3}{2} R \times ∆ T$
$∆ T$ $=$ $\frac{166.17 \times 2}{3 \times 8.314} K$
$= 13.34 K$
So the change in temperature is $13.34 K$

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Similar questions

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

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
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 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.
$O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} O (l), E^{o} = 1.23 V 2 H^{+} (a q) + 2 e^{-} \to H_{2} (g), E^{o} = 0.00 V$
Use $F = 96500 C m o l^{- 1}, R = 8.314 J m o l^{- 1} K^{- 1}$

Q. Given the half-cell reactions
i)

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

ii)

2 H^{+} (a q) + \frac{1}{2} O_{2} (g) + 2 e^{-} \to H_{2} O (l); E^{0} = + 1.23 V

E^{0}

for the reaction

F e (s) + 2 H^{+} (a q) + \frac{1}{2} O_{2} (g) \to F e^{2 +} (a q) + H_{2} O (l)

Q. Which of the following reaction will provide maximum voltage?
Given:

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

A g^{+} (a q) + e^{-} \to A g (s); E^{0} = 0.80 V

2 H^{+} (a q) + 2 e^{-} \to H_{2} (g); E^{0} = 0 V

N i^{2 +} (a q) + 2 e^{-} \to N i (s); E^{0} = - 0.25 V

The volume $V$ of a given mass of monatomic gas changes with temperature $T$ according to the relation $V = K T^{\frac{2}{3}}$ . The work done when temperature changes by $90 K$ will be $x R$ . The value of $x$ is ____. [ $R =$ universal gas constant]

Q. Which of the following metals will couple to provide a higher cell potential for a galvanic cell?

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

N i^{2 +} (a q) + 2 e^{-} \to N i (s); E^{0} = - 0.25 V

P b^{2 +} (a q) + 2 e^{-} \to P b (s); E^{0} = - 0.13 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

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