Question

Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 ∘C: 3Pb2+(aq)+2Cr(s)→3Pb(s)+2Cr3+(aq)(Note that the equation is balanced.)

Answer 1

Answer : The standard cell potential of the reaction is, -1.46 V

Explanation :

The given balanced cell reaction is,  

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

Here, chromium (Cr) undergoes oxidation by loss of electrons and act as an anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

The standard values of cell potentials are:

Standard reduction potential of lead $E^0_{[Pb^{2+}/Pb]}=-0.13V$

Standard reduction potential of chromium $E^0_{[Cr^{3+}/Cr]}=1.33V$

Now we have to calculate the standard cell potential for the following reaction.

$E^0=E^0_{cathode}-E^0_{anode}$

$E^0=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cr^{3+}/Cr]}$

$E^0=(-0.13V)-1.33V=-1.46V$

Therefore, the standard cell potential of the reaction is, -1.46 V