Calculate E_{cell} for the following cell:Pt_ | vedclass

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(A) The cell reaction is: $H_{2(g)} + 2H^{+}_{(cathode)} ightarrow 2H^{+}_{(anode)} + H_{2(g)}$For weak acid $HA$,$[H^{+}]_{anode} = \sqrt{K_a 	ime

Vedclass · Accepted Answer

$0.06$

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(A) The cell reaction is: $H_{2(g)} + 2H^{+}_{(cathode)} ightarrow 2H^{+}_{(anode)} + H_{2(g)}$For weak acid $HA$,$[H^{+}]_{anode} = \sqrt{K_a 	imes C} = \sqrt{10^{-7} 	imes 1} = 10^{-3.5} \ M$.For weak acid $HB$,$[H^{+}]_{cathode} = \sqrt{K_a 	imes C} = \sqrt{10^{-5} 	imes 1} = 10^{-2.5} \ M$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log Q$.Here,$E^{\circ}_{cell} = 0$,$n = 2$,and $Q = \frac{[H^{+}]^2_{anode}}{[H^{+}]^2_{cathode}} = \frac{(10^{-3.5})^2}{(10^{-2.5})^2} = \frac{10^{-7}}{10^{-5}} = 10^{-2}$.$E_{cell} = 0 - \frac{0.06}{2} \log(10^{-2}) = -0.03 	imes (-2) = 0.06 \ V$.

Calculate $E_{cell}$ for the following cell:
$Pt_{(s)} | H_{2(g)} | HA, 1 \ M \ (K_a = 10^{-7}) || HB, 1 \ M \ (K_a = 10^{-5}) | H_{2(g)} | Pt_{(s)}$ (in $V$)

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Calculate $E_{cell}$ for the following cell:$Pt_{(s)} | H_{2(g)} | HA, 1 \ M \ (K_a = 10^{-7}) || HB, 1 \ M \ (K_a = 10^{-5}) | H_{2(g)} | Pt_{(s)}$ (in $V$)