Question

The following questions are about the following cell: All(s) | Al^3+ || Pb^2+ | Pb(s) (a) Identify the anode and the cathode. (b) Write the balanced overall chemical equation. (c) Calculate the potential for the cell under standard conditions.

Answer 1

Here we have the cell as

$Al(s)|Al^{3+}||Pb^{2+}|Pb(s)$

a) Here anode is

$Al(s)|Al^{3+}$

and the cell reaction is written as

$Al(s)\rightarrow Al^{3+}+3e.................eqn(1)$

and here Cathode is

$Pb^{2+}|Pb(s)$

and the cell reaction is written as

$Pb^{2+}+2e\rightarrow Pb(s)...............eqn(2)$

b) To get the balanced overall reaction , we nned to multiply 2 with eqn(1) and 3 with eqn (2), we get

$2Al(s)\rightarrow 2Al^{3+}+6e.................eqn(1)$

And

$3Pb^{2+}+6e\rightarrow 3Pb(s)...............eqn(2)$

Now add eqn(1) and eqn (2), we get

$3Pb^{2+}+2Al(s)\rightarrow 3Pb(s)+2Al^{3+}$

C) For Anode

$\\Al(s)|Al^{3+}\quad \quad E^\circ_{(oxd)}=+1.66V\\ OR\\ Al^{3+}|Al(s)\quad \quad E^\circ_{(red)}=-1.66V\\$

For cathode

$Pb^{2+}|Pb(s)\quad\quad E^\circ_{(red)}=-0.13V$

Now we know that

$\\E^\circ_{(cell)}=E^\circ_{(cathode)}-E^\circ_{(anode)}\\ Here\\ E^\circ_{(cell)}=-0.13V-(-1.66)\\ E^\circ_{(cell)}=-0.13V+1.66V\\ E^\circ_{(cell)}=1.53V\\$

here the cell potential is 1.53V