The emf of a cell corresponding to the reaction- Zn - 2H-aq- - Zn2- 0-1 M - H2g 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 volt E_cell = E^∘_cell 0.05912log[Zn^2+][H_2][H^+]^2 0.28 = 0.76 0.05912log(0.1)× 1[H^+]^2 log0.1[H^+]^2 = 2× 0.480.0591 ...

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Redox Reactions and Couple

The emf of a ...

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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 $E_{H^{+} / H_{2}}^{\circ} = 0$ .

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Z n + 2 H^{+} (a q) \to Z n^{2 +} (0.1 M) + H_{2} (g) (1 a t m)

25^{\circ}

(E_{Z n^{2 +} / Z n}^{\circ} = - 0.76

E_{H^{+} / H_{2}}^{\circ} = 0)

Z n_{(s)} + 2 H_{(a q)}^{+} \to Z n^{2 +} (0.1 M) + H_{2 (g)}

1

0.28

15^{o}

E_{Z n^{2 +} / Z n}^{o} = - 0.76

E_{H^{+} / H_{2}}^{o} = 0

Z n (s) + 2 H^{+} (aq) \to Z n^{2 +} (0.1 M) + H_{2} (g, 1 bar)

0.28 V

25^{\circ} C

E_{Z n^{2 +} | Z n}^{\circ} = - 0.763 V

Z n + 2 H^{+} \to Z n^{2 +} (0.1 M) + H_{2} (1 a t m)

0.28

25^{\circ} C

E_{(Z n^{2 +} | Z n)}^{⊖} = - 0.76 V

Z n (s) + 2 H^{⨁} (a q) \to Z n^{2 +} (0.1 M) + H_{2} (g, 1 a t m)

0.82 V

25^{\circ} C

E_{Z n^{2 +} | Z n}^{⊝} = - 0.76 V

E_{H^{⊝} | H_{2}}^{⊝} = 0

63 \times 10 - x

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