The following diagram represents a $H^+(aq)/H_2(g)$ half cell for the reaction $$2H^+(aq) + 2e^- \rightleftharpoons H_2(g)$$ **a.** **i.** For this half cell, identify an appropriate material for electrode Z - VCE - SSCE Chemistry - Question 7 - 2007 - Paper 1

The standard operating temperature for the half cell is 25°C (or 298 K).

The stronger reductant is Cadmium (Cd). This conclusion is based on the observation that as the pH in half cell 1 increased, it indicates that $H^+$ ions are being reduced, and since Cadmium reduces $Cd^{2+}$ ions with a more negative standard reduction potential compared to the hydrogen half cell reaction, it serves as a stronger reductant.

To calculate the amount of $X^{2-}$ that reacted:

1. Initial concentration: 1.00 M.
2. Final concentration: 0.725 M.
3. Change in volume for $X^{2-}$: 100.0 mL.
4. Amount reacted:

$\text{mol of } X^{2-} = (1.00 - 0.725) \times 0.100 \, \text{L} = 0.0275 \, ext{mol}$

From Faraday's law, charge (Q) can be related to moles of electrons (n) using:

$n(e^-) = \frac{Q}{F}$

Where $F$ = Faraday's constant ($96500 \, C/mol$).

So, $n(e^-) = \frac{2654 \, C}{96500 \, C/mol} = 0.0274 \, mol$

Thus, the ratio is:

$n(X^{2-}) : n(e^-) = 0.0275 : 0.0274 \approx 1 : 1$

The oxidation state of the product ($X^{2-}$) in half cell 1 is +2.

The half reaction at the electrode of half cell 1 can be written as:

$X^{2-} \rightarrow X^{2+} + 2e^-$