EMFreduction of the half cell NaOH 0-001 M - H 2 1atm Pt at 298 K isA- -0-1773B- -0-6501C- ZeroD- 0-0591

The correct option is B 0.6501[OH^ ]= 10^ 3 [H^+ ]=10^ 11 H^+_(aq)+e^ → H_2(g) E_reduction=0 0.0591/1log 1/10^ 11 = 0.0591× 11 = 0.6501 ...

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Question

EMF(reduction) of the half cell $\frac{N a O H_{(0.001 M)}}{H_{2 (1 a t m)} (P t)}$ at 298 K is

-0.1773

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-0.6501

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Zero

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0.0591

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Z n | Z n^{2 +} | | C u^{2 +} | C u

298 K

[Z n^{2 +}] = 0.2 M

[C u^{2 +}] = 4.0 M

C u^{2 +} + 2 e^{-} \to C u E^{\circ} = + 0.34 V

Z n^{2 +} + 2 e^{-} \to Z n E^{\circ} = - 0.76 V

E = E^{\circ} - (0.0591 / n) (log Q)

P t (H_{2} 1 a t m) | C H_{3} C H_{2} C O H (0.15 M) | | 0.01 M N H_{4} O H | H_{2} (1 a t m) P t

K_{a}

C H_{3} C H_{2} C O O H = 1.4 \times 10^{- 5}

K_{b}

N H_{4} O H 1.8 \times 10^{- 5}

298 K

Z n^{2 +} + 2 e^{-} \to Z n

[Z n^{2 +}] = 0.1 M

E^{\circ} = - 0.76 v o l t

C u^{2 +} / C u

C u^{2 +}

100

298 K ?

[\frac{2.303 R T}{F} = 0.06]

C u^{2 +} (0.1 M) / C u

100

298 K

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