Home
Class 12
CHEMISTRY
The half reactions that occur in a lead ...

The half reactions that occur in a lead acid battery are:
`PbSO_(3)(s)+2e^(-)toPb(s)+SO_(4)^(2-)(aq)E^(@)=-0.36V`
`PbO_(2)(s)+4H^(+)(aq)+SO_(4)^(2-)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(l)E^(@)=+1.69V`
Calculate the overall potential for the cell in discharging reaction, `E_(cell)^(@)` Give answer in nearest single digit integer.

Text Solution

Verified by Experts

The correct Answer is:
2
Use the `E^(@)` values to decide the direction of the overall cell reaction.
the `E^(@)` value for the reduction of `PbO_(2)` is more positive than that for the reduction of `PbSO_(4)`, so the `PbO_(2)` half reaction will oxidize `Pb` to `PbSO_(4)`.
Anode `Pb(S)+SO_(4)^(2-(aq)toPbSO_(4)(s)+2e^(-)`
Cathode
`PbO_(2)(s)+4H(aq)+SO_(4)^(2-)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(I)`
`E_(cell)^(@)=(E_("cathode")^(@))_(RP)-(E_("anode")^(@))_(RP)`
`=(+1.69V)-(-0.36V)=+2.05V`
Promotional Banner

Similar Questions

Explore conceptually related problems

The reduction potential of the two half cell reaction (occurring in an electrochemical cell) are PbSO_(4)(s)+2e^(-)rarrPb(s)+SO_(4)^(2-)(aq)(E^(@)=-0.31V) Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=0.80V) The feasible reaction will be

Given the folowing standerd electrode potentials, the K_(sp) for PbBr^(2) is : PbBr_(2)(s)+2e^(-)toPb(s)+2Br^(-)(aq), E^(@)=-0.248 V Pb^(2+)(aq)+2e^(-)toPb(s), E^(@)=-0.126 V

The overall reaction for the lead storage batter when it discharges is: Pb(s) + PbO_(2)(s) + 4H^(+)(aq) + 2SO_94)^(2-)(aq) rightarrow 2PbSO_(4)(s) + 2H_(2)O(l) (P) PbSO_(4) is formed only at the cathode. (Q) The density of the solution decreases. Which statements correctly describes the battery as its discharges?

The reduction potential of the two half cell reactions (occuring in an electrochemical cell) are PbSO_(4)+ 2e^(-)rarrPb+SO_(4)^(2-) (E^(@)=-0.31V) Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=+0.80V) The fessible reaction will be

The half cell reactions with reduction potentials are Pb(s) to Pb^(+2) (aq), E_("red")^(@) =-0.13 V Ag(s) to Ag^(+) (aq) , E_("rod")^(@) =+0.80 V Calculate its emf.

In the equation below, which species acts the oxidation agent? Pb(s)+PbO_(2)(s)+H^(+)(aq)+2HSO_(4)^(-)(aq)rarr2PbSO_(4)(s)+2H_(2)O(l)

Consider the following half reactions : PbO_(2)(s)+4H^(o+)(aq)+SO_(4)^(2-)+2e^(-)rarrPbSO_(4)9s)+2H_(2)O" "E^(c-)=+1.70V PbSO_(4)(s)+2e^(-)rarr Pb(s)+SO_(4)^(2-)(aq)" "E^(c-)=-0.31V a. Calculate the value of E^(c-) for the cell. b. Calculate the voltage generated by the cell if [H^(o+)]=0.10M and [SO_(4)^(2-)]=2.0M c. What voltage is generated by the cell when it is at chemical equilibrium ?

If the half cell reactions are given as (i) Fe^(2+)(aq)+2e^(-)rarrFe(s),E^(@)=0.44V (ii) 2H^(+)(aq)+ 1//2O_(2)(g)+2e^(-)rarrH_(2)O(l),E^(@)=+1.23V The E^(@) for the reaction.