At 298 K,the emf of the cell is .......... V | vedclass

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(C) The cell reaction is:Anode: $H_{2(2 \ atm)} \rightarrow 2H^{+}_{(0.002 \ M)} + 2e^{-}$Cathode: $2H^{+}_{(0.1 \ M)} + 2e^{-} \rightarrow H_{2(1 \ a

Vedclass · Accepted Answer

$0.1$

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(C) The cell reaction is:Anode: $H_{2(2 \ atm)} ightarrow 2H^{+}_{(0.002 \ M)} + 2e^{-}$Cathode: $2H^{+}_{(0.1 \ M)} + 2e^{-} ightarrow H_{2(1 \ atm)}$Overall reaction: $H_{2(2 \ atm)} + 2H^{+}_{(0.1 \ M)} ightarrow 2H^{+}_{(0.002 \ M)} + H_{2(1 \ atm)}$Using the Nernst equation at $298 \ K$:$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$Since $E^{\circ}_{cell} = 0 \ V$ for a concentration cell and $n = 2$:$Q = \frac{[H^{+}]^2_{anode} 	imes P_{H_2(cathode)}}{[H^{+}]^2_{cathode} 	imes P_{H_2(anode)}} = \frac{(0.002)^2 	imes 1}{(0.1)^2 	imes 2} = \frac{4 	imes 10^{-6}}{0.01 	imes 2} = \frac{4 	imes 10^{-6}}{0.02} = 2 	imes 10^{-4}$$E_{cell} = 0 - \frac{0.0591}{2} \log(2 	imes 10^{-4})$$E_{cell} = -0.02955 	imes (\log 2 + \log 10^{-4})$$E_{cell} = -0.02955 	imes (0.301 - 4) = -0.02955 	imes (-3.699) \approx 0.109 \ V$Given the options,the closest value is $0.1 \ V$.

At $298 \ K$,the $emf$ of the cell is .......... $V$.
$Pt | H_{2(2 \ atm)} | H^{+}_{(0.002 \ M)} || H^{+}_{(0.1 \ M)} | H_{2(1 \ atm)} | Pt$

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At $298 \ K$,the $emf$ of the cell is .......... $V$.$Pt | H_{2(2 \ atm)} | H^{+}_{(0.002 \ M)} || H^{+}_{(0.1 \ M)} | H_{2(1 \ atm)} | Pt$