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Standard XII

Chemistry

EMF

Calculate the...

Question

Calculate the emf of the following cell:
$P t (H_{2} 1 a t m) | C H_{3} C H_{2} C O H (0.15 M) | | 0.01 M N H_{4} O H | H_{2} (1 a t m) P t$
$K_{a}$ for $C H_{3} C H_{2} C O O H = 1.4 \times 10^{- 5}$
$K_{b}$ for $N H_{4} O H 1.8 \times 10^{- 5}$ .

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Solution

$[H^{+}]$ in $C H_{3} C H_{2} C O O H = \sqrt{C \times K_{a}} = \sqrt{0.15 \times 1.4 \times 10^{- 5}}$
$= 1.449 \times 10^{- 3}$
$[O H^{-}]$ in $N H_{4} O H = \sqrt{C \times K_{b}} = \sqrt{0.01 \times 1.8 \times 10^{- 5}} = 0.4242 \times 10^{- 3}$
$[H^{+}]$ in $N H_{4} O H = \frac{10^{- 14}}{0.4242 \times 10^{- 3}} = 2.3573 \times 10^{- 11}$
$E_{c e l l} = 0.0591 log \frac{[H^{+}]_{R H S}}{[H^{+}]_{L H S}}$
$- 0.4603 v o l t$ .

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Similar questions

Calculate the emf of the cell

Pt, $H_{2} (1.0 a t m)$ | $C H_{3} C O O H$ (0.1M) || $N H_{3}$ (aq, 0.001M) | $H_{2}$ (1.0 atm), Pt Ka( $C H_{3} C O O H$ ) = $10^{- 5}$ , Kb(NH3) = $10^{- 5}$

Calculate the e.m.f. of cell in Volts.

\begin{matrix} P t_{H_{2}} 1 a t m \end{matrix} ∣ ∣ ∣ \begin{matrix} C H_{3} C O O H 0.1 M \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} N H_{4} O H 0.01 M \end{matrix} ∣ ∣ ∣ \begin{matrix} P t_{H_{2}} 1 a t m \end{matrix}

K_{a}

for

C H_{3} C O O H = 1.8 \times 10^{- 5}

;

K_{b}

for

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

If answer is

x

, then write as nearest integer to

10 | x |

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Q. Neglecting the liquid-liquid junction potential, calculate the emf of the following cell at

25^{\circ} C

H_{2} (1 a t m) | 0.4 M H C O O H ∥ 1 M C H_{3} C O O H | (1 a t m) H_{2}

K_{a}

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H C O O H

and

C H_{3} C O O H

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and

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Q. A Student planned an experiment that would use

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Calculate the emf of the following cell:Pt(H2 1 atm)|CH3CH2COH(0.15 M)||0.01 M NH4OH|H2(1 atm)PtKa for CH3CH2COOH=1.4×10−5Kb for NH4OH1.8×10−5.

[H+] in CH3CH2COOH=√C×Ka=√0.15×1.4×10−5=1.449×10−3[OH−] in NH4OH=√C×Kb=√0.01×1.8×10−5=0.4242×10−3[H+] in NH4OH=10−140.4242×10−3=2.3573×10−11Ecell=0.0591log[H+]RHS[H+]LHS−0.4603 volt.

Calculate the emf of the following cell:
$P t (H_{2} 1 a t m) | C H_{3} C H_{2} C O H (0.15 M) | | 0.01 M N H_{4} O H | H_{2} (1 a t m) P t$
$K_{a}$ for $C H_{3} C H_{2} C O O H = 1.4 \times 10^{- 5}$
$K_{b}$ for $N H_{4} O H 1.8 \times 10^{- 5}$ .