What will be the EMF for the given cell?Pt | | vedclass

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(B) The cell reaction is: $H_2(P_1) \rightarrow 2H^{+} + 2e^{-}$ (Anode) and $2H^{+} + 2e^{-} \rightarrow H_2(P_2)$ (Cathode).Overall reaction: $H_2(P

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$\frac{RT}{2F} \ln \frac{P_1}{P_2}$

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(B) The cell reaction is: $H_2(P_1) \rightarrow 2H^{+} + 2e^{-}$ (Anode) and $2H^{+} + 2e^{-} \rightarrow H_2(P_2)$ (Cathode).Overall reaction: $H_2(P_1) \rightarrow H_2(P_2)$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln Q$.Here,$E^{\circ}_{cell} = 0 \ V$,$n = 2$,and $Q = \frac{P_2}{P_1}$.$E_{cell} = 0 - \frac{RT}{2F} \ln \frac{P_2}{P_1}$.$E_{cell} = \frac{RT}{2F} \ln \frac{P_1}{P_2}$.

What will be the $EMF$ for the given cell?
$Pt | H_2(g, P_1) | H^{+}(aq.) (1 \ M) || H^{+}(aq.) (1 \ M) | H_2(g, P_2) | Pt$

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What will be the $EMF$ for the given cell?$Pt | H_2(g, P_1) | H^{+}(aq.) (1 \ M) || H^{+}(aq.) (1 \ M) | H_2(g, P_2) | Pt$