The emf of a cell corresponding to the reaction,
$Z n + 2 H^{+} (a q .) \to Z n^{2 +} (0.1 M) + H_{2} (g) 1 a t m$
is $0.28$ volt at $25^{\circ} C$ . Write the half-cell reactions and calculate the $p H$ of the solution at the hydrogen electrode.
$E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 v o l t$ and $E_{H^{+} / H_{2}}^{\circ} = 0$ .

Question

The emf of a cell corresponding to the reaction,
$Z n + 2 H^{+} (a q .) \to Z n^{2 +} (0.1 M) + H_{2} (g) 1 a t m$
is $0.28$ volt at $25^{\circ} C$ . Write the half-cell reactions and calculate the $p H$ of the solution at the hydrogen electrode.
$E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 v o l t$ and $E_{H^{+} / H_{2}}^{\circ} = 0$ .

Toppr Admin · Accepted Answer

E-x2218-cell-0-76-xA0-voltEcell-E-x2218-cell-x2212-0-05912log-Zn2-H2-H-20-28-0-76-x2212-0-05912log-0-1-xD7-1-H-2log0-1-H-2-2-xD7-0-480-0591log0-1-x2212-log-H-2-16-2436 -Since- -x2212-log-H-pH-2-xA0-pH-16-2436-x2212-log0-1pH-17-24362-8-6218-

The emf of a cell corresponding to the reaction,Zn+2H+(aq.)→Zn2+(0.1 M)+H2(g)1 atmis 0.28 volt at 25∘C. Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode.E∘Zn2+/Zn=−0.76 volt and E∘H+/H2=0.

E∘cell=0.76 voltEcell=E∘cell−0.05912log[Zn2+][H2][H+]20.28=0.76−0.05912log(0.1)×1[H+]2log0.1[H+]2=2×0.480.0591log0.1−log[H+]2=16.2436 [Since, −log[H+]=pH]2 pH=16.2436−log0.1pH=17.24362=8.6218.