State the formula of hypophosphorous acid.
$\text{H}_2\text{PO}_2^{-}$ is a strong reducing agent. It can be used to reduce metal cations without electrolysis. Equation 1: $\text{HPO}_3^{2-} + 2\text{H}_2\text{O} + 2e^{-} \rightleftharpoons \text{H}_2\text{PO}_2^{-} + 3\text{OH}^-$ $E^{\circ} = -1.57\,\text{V}$ In an experiment, an alkaline $\text{HPO}_3^{2-}/\text{H}_2\text{PO}_2^{-}$ half-cell is set up with $[\text{H}_2\text{PO}_2^{-}] = 0.050\,\text{mol dm}^{-3}$. All the other ions are at their standard concentration. Predict how the value of $E$ for this half-cell differs from its $E^{\circ}$ value. Explain your answer.
The $\text{Cr}^{3+}/\text{Cr}$ half-cell has a standard electrode potential of $-0.74\,\text{V}$. An electrochemical cell consists of an alkaline $\text{HPO}_3^{2-}/\text{H}_2\text{PO}_2^{-}$ half-cell and a $\text{Cr}^{3+}/\text{Cr}$ half-cell. Calculate the standard cell potential, $E^{\circ}_{\text{cell}}$.
Complete Fig. 2.1 so that it shows how the standard electrode potential of the $\text{Cr}^{3+}/\text{Cr}$ half-cell is measured relative to the standard hydrogen electrode. Name the chemicals, conditions and the apparatus that are needed.
Label Fig. 2.1 to show: • which electrode is positive • the direction in which electrons flow in the external circuit.
$\text{H}_2\text{PO}_2^{-}$ reduces $\text{Ni}^{2+}$ to Ni in alkaline conditions. Use equation 1 to build the ionic equation for this reaction. Equation 1: $\text{HPO}_3^{2-} + 2\text{H}_2\text{O} + 2e^{-} \rightleftharpoons \text{H}_2\text{PO}_2^{-} + 3\text{OH}^-$
$\text{H}_2\text{PO}_2^{-}(aq)$ reacts with $\text{OH}^-(aq)$. $\text{H}_2\text{PO}_2^{-}(aq) + \text{OH}^-(aq) \rightarrow \text{HPO}_3^{2-}(aq) + \text{H}_2(g)$ The volume of $\text{H}_2$ was recorded under room conditions. Use the molar volume of gas, $V_m$, together with the data from experiment 1, to calculate the rate of reaction in $\text{mol dm}^{-3}\text{ s}^{-1}$.
The rate equation has been found to be: $\text{rate} = k[\text{H}_2\text{PO}_2^{-}(aq)][\text{OH}^-(aq)]^2$ Show that the data in Table 2.1 agree with the rate equation.
State the units of the rate constant, $k$, for this reaction.
The experiment is repeated with a large excess of $\text{OH}^-(aq)$. Under these conditions, the rate equation becomes: rate = $k_1 [\text{H}_2\text{PO}_2^-(aq)]$ $k_1 = 8.25 \times 10^{-5}\text{ s}^{-1}$ Calculate the half-life, $t_{\frac{1}{2}}$, of the reaction.
Describe the effect that an increase in temperature has on the rate constant, $k_1$.
A student thinks that the reaction between $\text{H}_2\text{PO}_2^-(aq)$ and $\text{OH}^-(aq)$ could proceed faster if a heterogeneous catalyst were present. Describe how a heterogeneous catalyst works.