Question

2. Calculate cell potential for the following cell:
Pb/Pb2+(1 x 10-3M) || Pt, Cl, (1 bar)/Cl- (0.1 M)
Given that E
Pb2+ /Pb
0.13 V and
E
3C12/C1 = 1.358 V.​

Answer 1

1.488 V is the potential of the given cell.

Explanation:

Given that,

Cell reaction is as follows:

Pb + $2Cl^{-}$ → $Pb^{+2}$ + $Cl_{2}$

and

$E^{0} = Pb^{+2} /Pb = -0.13 V$

$E^{0} = \frac{1}{2} Cl_{2}/Cl^{-}$ = 1.358 V

So,

The potential of the cell:

$E^{0}_{cell}$ = $E^{0}$/(RHS) - $E^{0}(LHS)$

= 1.358 - (- 0.13V)

= 1.488 V

Learn more: Potential cell

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