Which one of the following metals will not reduce H2O?

The correct answer is Option (3) → Cu

• A metal reduces water when it is oxidised and sets free hydrogen gas. This is judged from its standard electrode potential (E°).

• A metal with a negative E° is a stronger reducing agent than hydrogen; such a metal can set free \( \ce{H2} \) from water. A metal with a positive E° cannot set free \( \ce{H2} \) even from acids, so it cannot reduce water.

The values for the four metals are:

    \( \displaystyle E^{\circ}_{\ce{Li^+/Li}}=-3.05\,\mathrm{V} \quad\quad E^{\circ}_{\ce{Ca^2+/Ca}}=-2.87\,\mathrm{V} \)
    \( \displaystyle E^{\circ}_{\ce{Fe^2+/Fe}}=-0.44\,\mathrm{V} \quad\quad E^{\circ}_{\ce{Cu^2+/Cu}}=+0.34\,\mathrm{V} \)

Only copper has a positive E°, so it is the one metal here that does not reduce water.

Option (1) Li — Reduces water. Lithium has the most negative E° of all (−3.05 V), so it is the most powerful reducing agent in aqueous solution and sets free hydrogen from water:

    \( \displaystyle \ce{2Li + 2H2O -> 2LiOH + H2 ^} \)

Option (2) Fe — Reduces water. Iron has a negative E° (−0.44 V), so it is a stronger reducing agent than hydrogen, though a much weaker one than Li or Ca.

Option (3) Cu — Does not reduce water. Copper has a positive E° (+0.34 V): it is a weaker reducing agent than hydrogen, so it cannot liberate \( \ce{H2} \) from water or even from dilute acids.

Option (4) Ca — Reduces water. Calcium has a large negative E° (−2.87 V), so it is a strong reducing agent and sets free hydrogen from water:

    \( \displaystyle \ce{Ca + 2H2O -> Ca(OH)2 + H2 ^} \)

NCERT (2026-27 edition):
Table 2.1: Standard Electrode Potentials at 298 K – “… 2. A positive E^o means that the redox couple is a weaker reducing agent than the H⁺/H₂ couple.”

Trends in the M²⁺/M Standard Electrode Potentials – “The unique behaviour of Cu, having a positive E°, accounts for its inability to liberate H₂ from acids.”

Table 2.1: Standard Electrode Potentials at 298 K