Consider the electrochemical cell: Pt | O_{ | vedclass

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(B) The cell reaction is $M^{2+} + H_2O \rightarrow M + \frac{1}{2}O_2 + 2H^+$. For the reaction to be spontaneous, $E_{cell} > 0$. At the limiting co

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The cell reaction is $M^{2+} + H_2O ightarrow M + \frac{1}{2}O_2 + 2H^+$. For the reaction to be spontaneous, $E_{cell} > 0$. At the limiting condition, $E_{cell} = 0$, so $E_{cathode} = E_{anode}$. $E_{cathode} = E^{\circ}_{M^{2+}/M} - \frac{0.059}{2} \log \frac{1}{[M^{2+}]} = 0.994 - 0 = 0.994 \ V$. $E_{anode} = E^{\circ}_{O_2/H_2O} - \frac{0.059}{4} \log \frac{1}{[H^+]^4 P_{O_2}^{1/2}} = 1.23 + \frac{0.059}{4} \log ([H^+]^4 	imes 1) = 1.23 + 0.059 \log [H^+] = 1.23 - 0.059 	imes pH$. Equating $E_{cathode} = E_{anode}$: $0.994 = 1.23 - 0.059 	imes pH$. $0.059 	imes pH = 1.23 - 0.994 = 0.236$. $pH = \frac{0.236}{0.059} = 4$. Thus, the nearest integer is $4$.

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