Consider the cell:Pt | H_2(P_1 atm) | H^{+}( | vedclass

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(C) The cell reaction is: $H_2(P_1) + 2H^{+}(x_2) \rightarrow 2H^{+}(x_1) + H_2(P_2)$.Using the Nernst equation for the cell potential $E_{cell} = E^0

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$x_1 = x_2$ and $P_1 > P_2$

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(C) The cell reaction is: $H_2(P_1) + 2H^{+}(x_2) \rightarrow 2H^{+}(x_1) + H_2(P_2)$.Using the Nernst equation for the cell potential $E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \frac{x_1^2 \cdot P_2}{x_2^2 \cdot P_1}$.Since $E^0_{cell} = 0$,$E_{cell} = -0.0591 \log \left( \frac{x_1}{x_2} \right) \left( \frac{P_2}{P_1} \right)^{1/2}$.For the reaction to be spontaneous,$E_{cell} > 0$.If $x_1 = x_2$,then $E_{cell} = -0.0591 \log \left( \frac{P_2}{P_1} \right)^{1/2} = -0.02955 \log \left( \frac{P_2}{P_1} \right) = 0.02955 \log \left( \frac{P_1}{P_2} \right)$.For $E_{cell} > 0$,we require $\log \left( \frac{P_1}{P_2} \right) > 0$,which implies $P_1 > P_2$.

Consider the cell:
$Pt | H_2(P_1 \ atm) | H^{+}(x_1 \ M) || H^{+}(x_2 \ M) | H_2(P_2 \ atm) | Pt$.
The cell reaction will be spontaneous if

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Consider the cell:$Pt | H_2(P_1 \ atm) | H^{+}(x_1 \ M) || H^{+}(x_2 \ M) | H_2(P_2 \ atm) | Pt$.The cell reaction will be spontaneous if