For the electrochemical cell shown below:Pt m | vedclass

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(C) The cell reaction is: $H_{2}(g) + Cu^{2+}(aq) \rightarrow 2H^{+}(aq) + Cu(s)$.Here,$n = 2$.The standard cell potential is $E^{\circ}_{cell} = E^{\

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(C) The cell reaction is: $H_{2}(g) + Cu^{2+}(aq) ightarrow 2H^{+}(aq) + Cu(s)$.Here,$n = 2$.The standard cell potential is $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 \, V - 0 \, V = 0.34 \, V$.Using the Nernst equation at $298 \, K$:$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[H^{+}]^{2}}{P_{H_{2}} [Cu^{2+}]}$$0.49 = 0.34 - \frac{0.0591}{2} \log \frac{x^{2}}{1 	imes 1}$$0.15 = -\frac{0.0591}{2} 	imes 2 \log x$$0.15 = -0.0591 	imes \log x$$\log x = -\frac{0.15}{0.0591} \approx -2.538$Since $pH = -\log [H^{+}] = -\log x$,we get $pH \approx 2.54$.The closest value is $2.5$.

Similar Questions

Calculate the equilibrium constant of the reaction,$Cu_{(s)} + 2 Ag^{+}_{(aq)} \longrightarrow Cu^{2+}_{(aq)} + 2 Ag_{(s)}$,given that for the reaction $E^{\circ}_{cell} = 0.46 \ V$.

Represent the cell in which the following reaction takes place:
$Mg_{(s)} + 2Ag^{+}(0.0001 \, M) \rightarrow Mg^{2+}(0.130 \, M) + 2Ag_{(s)}$
Calculate its $E_{cell}$ if $E^{\Theta}_{cell} = 3.17 \, V$.

If $E^{\circ}(Mg^{+2}_{(aq)} \mid Mg_{(s)}) = -2.37 \ V$. What is the potential for $Mg_{(s)} \rightarrow Mg^{+2}_{(0.01 \ M)} + 2 \overline{e}$ at $298 \ K$?

Write the Nernst equation for the $E_{cell}$ of a Daniell cell.

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