E^o_{cell} for the cell: Pt_{(s)} | H_{2(g)} | vedclass

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(B) The cell reaction is: $CH_3COOH + HCOO^- \rightleftharpoons CH_3COO^- + HCOOH$.The equilibrium constant for this reaction is $K_{eq} = \frac{K_a(C

Vedclass · Accepted Answer

$-0.0672$

Vedclass · Answer

(B) The cell reaction is: $CH_3COOH + HCOO^- ightleftharpoons CH_3COO^- + HCOOH$.The equilibrium constant for this reaction is $K_{eq} = \frac{K_a(CH_3COOH)}{K_a(HCOOH)} = \frac{1.8 	imes 10^{-5}}{2.4 	imes 10^{-4}} = 0.075$.Using the relation $E^o_{cell} = \frac{0.059}{n} \log K_{eq}$ at $25^oC$ (where $n=1$):$E^o_{cell} = 0.06 	imes \log(0.075) = 0.06 	imes \log(\frac{3}{40}) = 0.06 	imes (\log 3 - \log 40) = 0.06 	imes (0.477 - 1.602) = 0.06 	imes (-1.125) = -0.0675 \ V$.Given the approximation $\frac{2.303RT}{F} = 0.06$,the value is approximately $-0.0672 \ V$.

$E^o_{cell}$ for the cell: $Pt_{(s)} | H_{2(g)} | HCOOH_{(aq)} || CH_3COOH_{(aq)} | H_{2(g)} | Pt_{(s)}$ at $25^oC$ is ............ $V$. ($K_a$ of $HCOOH = 2.4 \times 10^{-4}$,$K_a$ of $CH_3COOH = 1.8 \times 10^{-5}$,$\log 2 = 0.3$,$\log 3 = 0.477$,$\frac{2.303RT}{F} = 0.059$ or $0.06$)

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