CAIE A-Level Chemistry A2 24.2.8 How the Value of an Electrode Potential E Topic Practice

Hypophosphorous acid is an inorganic acid.
The conjugate base of hypophosphorous acid is $\mathrm{H}_{2} \mathrm{PO}_{2}^{-}$.

$\mathrm{H}_{2} \mathrm{PO}_{2}^{-}$is a strong reducing agent. It can be used to reduce metal cations without the need for electrolysis.
equation $1 \quad \mathrm{HPO}_{3}^{2-}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{2}^{-}+3 \mathrm{OH}^{-}$$E^{\ominus}=-1.57 \mathrm{~V}$

In an experiment, an alkaline $\mathrm{HPO}_{3}{ }^{2-} / \mathrm{H}_{2} \mathrm{PO}_{2}{ }^{-}$half-cell is constructed with $\left[\mathrm{H}_{2} \mathrm{PO}_{2}^{-}\right]=0.050 \mathrm{~mol} \mathrm{dm}^{-3}$.

All other ions are at their standard concentration.

Predict how the value of E of this half-cell differs from its $E^{\ominus}$ value.

Explain your answer.