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)/(2)O_(2)(g)rarr H_(2)O(l)`.
The work derived from the cell on the consumption of `1.0xx10^(-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 gien below.
`O_(2)(g)+4H^(+)(aq)+4e^(-)rarr 2H_(2)O(l), E^(@)=1.23V`,
`2H^(+)(aq)+2e^(-)rarrH_(2)(g), E^(@)=0.00V.`
Use `F="96500 C mol"^(-1),="8.314 J mol"^(-1)K^(-1)`.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the standard cell potential (E°) of the fuel cell reaction.
The cell reaction is:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \]

Using the given standard reduction potentials:
- For the reduction of oxygen:
\[ \text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \rightarrow 2\text{H}_2\text{O}(l), \quad E^\circ = 1.23 \, \text{V} \]
- For the oxidation of hydrogen:
\[ 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g), \quad E^\circ = 0.00 \, \text{V} \]

The overall cell potential (E°) is given by:
\[ E^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.23 \, \text{V} - 0.00 \, \text{V} = 1.23 \, \text{V} \]

### Step 2: Calculate the standard Gibbs free energy change (ΔG°) for the reaction.
Using the formula:
\[ \Delta G^\circ = -nFE^\circ \]
Where:
- \( n = 2 \) (number of moles of electrons transferred)
- \( F = 96500 \, \text{C/mol} \)
- \( E^\circ = 1.23 \, \text{V} \)

Calculating ΔG°:
\[ \Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times 1.23 \, \text{V} \]
\[ \Delta G^\circ = -237390 \, \text{J/mol} \]

### Step 3: Calculate the work done (W) by the fuel cell on the consumption of \( 1.0 \times 10^{-3} \) mol of \( \text{H}_2(g) \).
Given the efficiency of the fuel cell is 70%, the work done can be calculated as:
\[ W = \text{Efficiency} \times \Delta G^\circ \times \text{moles of } H_2 \]
\[ W = 0.70 \times (-237390 \, \text{J/mol}) \times (1.0 \times 10^{-3} \, \text{mol}) \]
\[ W = 0.70 \times 237.390 \, \text{J} \]
\[ W \approx 166.173 \, \text{J} \]

### Step 4: Use the work done to find the change in temperature (ΔT) of the monoatomic ideal gas.
Using the adiabatic process formula:
\[ W = nC_v\Delta T \]
For a monoatomic ideal gas, \( C_v = \frac{3}{2}R \) and \( n = 1 \, \text{mol} \):
\[ W = 1 \times \frac{3}{2}R\Delta T \]
Substituting \( R = 8.314 \, \text{J/(mol K)} \):
\[ W = \frac{3}{2} \times 8.314 \Delta T \]
Now substituting the value of \( W \):
\[ 166.173 = \frac{3}{2} \times 8.314 \Delta T \]

### Step 5: Solve for ΔT.
Rearranging gives:
\[ \Delta T = \frac{166.173}{\frac{3}{2} \times 8.314} \]
\[ \Delta T = \frac{166.173}{12.471} \]
\[ \Delta T \approx 13.32 \, \text{K} \]

### Final Answer:
The change in temperature of the ideal gas is approximately \( 13.32 \, \text{K} \).

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