Question

Balance the following skeleton reaction, calculate E o cell , and state whether the reaction is spontaneous: Cu+(aq) + PbO2(s) + SO42−(aq) → PbSO4(s) + Cu2+(aq) [acidic] Write the balanced equation, making sure to include the state of each component. E o cell = V Is the reaction spontaneous or nonspontaneous? spontaneous nonspontaneous

Answer 1

Answer:

$2Cu^++PbO_2+SO_4^{2-}+4H^+\rightarrow 2Cu^{2+}+PbSO_4+2H_2O$

$E^0=1.53$, the reaction is spontaneous.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$Cu^++PbO_2+SO_4^{2-}\rightarrow PbSO_4+Cu^{2+}$

The half cell reactions for the above reaction follows:

Oxidation half reaction:  $Cu^+\rightarrow Cu^{2+}+e^-$

Reduction half reaction:  $PbO_2+SO_4^{2-}+4H^++2e^-\rightarrow PbSO_4+2H_2O$

To balance the oxidation half reaction must be multiplied by 2 and thus, the balanced equation is:-

$2Cu^++PbO_2+SO_4^{2-}+4H^++2e^-\rightarrow 2Cu^{2+}+PbSO_4+2H_2O$

Here $Cu^+$  undergoes oxidation by loss of electrons, thus act as anode. $PbO_2$  undergoes reduction by gain of electrons and thus act as cathode.

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

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu^{+}]}= +0.16V$

$E^0_{[PbO_2/PbSO_4]}=+1.69V$

$E^0=E^0_{[PbO_2/PbSO_4]}- E^0_{[Cu^{2+}/Cu^{+}]}$

$E^0=+1.69- (0.16V)=1.53$

Since, $E^0>0$, the reaction is spontaneous.