The E^0_{cell} of the cell Al(s)|Al^{3+}(1M)| | vedclass

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Question

(B) The cell reaction is: $2Al(s) + 3Pb^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Pb(s)$.The standard cell potential is given by: $E^0_{cell} = E^0_{cathod

Vedclass · Accepted Answer

$-1.64$

Vedclass · Answer

(B) The cell reaction is: $2Al(s) + 3Pb^{2+}(aq) ightarrow 2Al^{3+}(aq) + 3Pb(s)$.The standard cell potential is given by: $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.Here, the cathode is $Pb^{2+}/Pb$ and the anode is $Al^{3+}/Al$.So, $E^0_{cell} = E^0_{Pb^{2+}/Pb} - E^0_{Al^{3+}/Al}$.Given $E^0_{cell} = 1.5 	ext{ V}$ and $E^0_{Pb^{2+}/Pb} = -0.14 	ext{ V}$.Substituting the values: $1.5 	ext{ V} = -0.14 	ext{ V} - E^0_{Al^{3+}/Al}$.$E^0_{Al^{3+}/Al} = -0.14 	ext{ V} - 1.5 	ext{ V} = -1.64 	ext{ V}$.

The $E^0_{cell}$ of the cell $Al(s)|Al^{3+}(1M)||Pb^{2+}(1M)|Pb(s)$ is $1.5 \text{ V}$. If $E^0_{Pb^{2+}/Pb}$ is $-0.14 \text{ V}$, then the standard electrode potential $E^0_{Al^{3+}/Al}$ will be: (in $\text{ V}$)

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