The following diagram represents a H⁺(aq)/H₂(g) half cell for the reaction 2H⁺(aq) + 2e⁻ ⇌ H₂(g) --- a - VCE - SSCE Chemistry - Question 7 - 2007 - Paper 1

The temperature at which this half cell must operate to be a standard half cell is 25°C.

The stronger reductant is Cd. This conclusion is reached by noting that as the pH in half cell 1 increased, it indicates that H⁺ ions have been consumed, therefore Cd must be reducing H⁺ to H₂, demonstrating that Cd is a stronger reductant compared to H⁺.

To calculate the amount of X²⁻ that reacted, we use the formula:

$ext{mol of } X^{2-} ext{ used} = (1.00 - 0.725) imes 0.1 = 0.0275 ext{ mol}$

The ratio of n(X²⁻) reacted to n(e⁻) can be calculated as follows:

• The total charge discharged is 2654 C.
• The moles of electrons (n(e⁻)) can be found using Faraday's law:

$n(e^{-}) = \frac{Q}{F} = \frac{2654 ext{ C}}{96500 ext{ C/mol}} = 0.0274 ext{ mol}$

Therefore, the ratio is:

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

The oxidation state of the product (X) in the half reaction in half cell 1 is +3.

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

$X^{2-} \rightarrow X^{3+} + 2e^{-}$