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}$.

Similar Questions

Which graph correctly correlates $E_{cell}$ as a function of concentrations for the cell (for different values of $M$ and $M'$):-
$Zn_{(s)} + Cu^{2+}(M) \to Zn^{2+}(M') + Cu_{(s)};$ $E^o_{cell} = 1.10 \, V$
$X$-axis : $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$Y$-axis : $E_{cell}$

Which of the following is correct as a Nernst equation for the given electrochemical cell ?
$Mg_{(s)}|Mg_{(aq)}^{2+}(0.1 \ M)||Cl_{(aq)}^{-}(0.1 \ M)|Cl_{2_{(g)}}(1 \ bar)|Pt_{(s)}$

If $Zn/Zn^{+2}$ electrode is diluted $100$ times,then the change in electromotive force will be:

At $298 \ K$,if the $emf$ of the cell corresponding to the reaction,$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$ is $0.28 \ V$,then the $pH$ of the solution at the hydrogen electrode is (Given: $\frac{2.303 \ RT}{F} = 0.06 \ V$,$E^o_{Zn^{2+}|Zn} = -0.76 \ V$)

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