A hydrogen electrode is prepared by placing into a solution of pH = 3. The magnitude of electrode potential at 25ºC is given by t × 10^–2 V. Find out t (Nearest integer)

The correct answer is 18.

Let us calculate the electrode potential of a hydrogen electrode under the given conditions.
A standard hydrogen electrode operates based on the reduction half-reaction:
${\mathrm{2H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{2e}}^{-}\to H_2\left(g\right)$

For a hydrogen electrode at ${\mathrm{25}}^{\circ }C$ (298 K) and under standard pressure of $H_2$ gas (1 bar or 1 atm), the Nernst equation for the reduction potential is:
$E=E^{\circ }-\frac{2.303RT}{F}log\frac{1}{\left[H^{+}\right]}$

Since the standard electrode potential of a hydrogen electrode ($E^{\circ }$) is defined as 0 V, and the value of $\frac{2.303RT}{F}$ at 298 K is approximately 0.0591 V, the equation simplifies to:
$E=-0.0591log\left({\left[H^{+}\right]}^{-1}\right)=0.0591log\left[H^{+}\right]$

By definition, $\mathrm{pH}=-log\left[H^{+}\right]$, which gives:
$E=-0.0591\times \mathrm{pH}$

Given that the pH of the solution is 3:
$E=-0.0591\times 3=-0.1773\text{V}$

The magnitude (absolute value) of this electrode potential is:
$\left|E\right|=0.1773\text{V}\approx 0.18\text{V}$

We are given that the magnitude of the electrode potential is expressed in the form $t\times {10}^{-2}\text{V}$:
$0.18\text{V}=18\times {10}^{-2}\text{V}$

Comparing the two values, we find:
$t=18$