The photoelectric current from Na (work funct | vedclass

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Question

(C) The cell reaction is:$\frac{1}{2} H_{2}(g) + AgCl_{(s)} \rightarrow H^{+}_{(aq)} + Ag_{(s)} + Cl^{-}_{(aq)}$The cell potential $E_{cell}$ is given

Vedclass · Accepted Answer

$142$

Vedclass · Answer

(C) The cell reaction is:$\frac{1}{2} H_{2}(g) + AgCl_{(s)} ightarrow H^{+}_{(aq)} + Ag_{(s)} + Cl^{-}_{(aq)}$The cell potential $E_{cell}$ is given by the Nernst equation:$E_{cell} = E^{0}_{cell} - 0.06 \log([H^{+}][Cl^{-}]) = 0.22 - 0.06 \log(10^{-1} 	imes 10^{-1}) = 0.22 + 0.12 = 0.34 \ V$For $Na$,the energy of the incident photon $h
u$ is:$h
u = w_{0}(Na) + E_{cell} = 2.3 \ eV + 0.34 \ eV = 2.64 \ eV$For $K$,the stopping potential $V_{s}$ required is:$V_{s} = h
u - w_{0}(K) = 2.64 \ eV - 2.25 \ eV = 0.39 \ V$Using the Nernst equation for the cell with unknown $pH$ $([H^{+}] = [Cl^{-}] = 10^{-pH})$:$0.39 = 0.22 - 0.06 \log([H^{+}]^{2}) = 0.22 - 0.06 	imes 2 \log[H^{+}] = 0.22 + 0.12 \ pH$$0.12 \ pH = 0.39 - 0.22 = 0.17$$pH = \frac{0.17}{0.12} = 1.4166 \approx 1.42$Thus,$pH = 142 	imes 10^{-2}$.

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