Home
Class 12
CHEMISTRY
Determining the directiom of spontaneity...

Determining the directiom of spontaneity from electrode pot`Na^(+)(aq.) + e^(-) rarr Na(s) E^(@) = -2.71 V`entials : Predict from Table 3.1, whether `Pb^(2+)(aq.)` can oxidize `Al(s)` or `Cu(s)` under standard conditions. Calculate `E^(ɵ)` for each reaction at `25^(@)C`.
Strategy: To predict whether a redox reaction is spontaneous, remember that an oxidizing agent can oxidize any reducing agent that lies below it in the table but can't oxidize one that lies above it.
Alternatively, write the expected reaction. Find the oxidizing agents in the equations, one is on the left side and the other on the right side. Locate these oxidizing agents in a table of electrode potentials (the oxidizing agent is on the left side of the reduction half-reaction). The stronger oxidizing agent is the one involved in the half-reaction with the more positive standard electrode potential.

Text Solution

Verified by Experts

`Pb^(2+)(aq.)` is above `Al(s)` in the table but below `(Cu)(s)`. Therefore, `Pb^(2+)(aq.)` can oxidize `Al(s)` but can't oxidize `Cu(s)`. To confirm these predictions, we can calculate `E^(@)` values for the overall reactions.
`{:(3xx[Pb^(2+)(aq.)2e^(-)hArr Pb(s)]" "E^(@)=-0.13V),(2xx[Al(s)hArr Al^(3+)(aq.)+3e]" "E^(@)=1.66V),(bar(3Pb^(2+)(aq.)+2Al(s)hArr3Pb(s)+2Al^(3+)(aq.)E^(@)=1.53V)):}`
For the oxidation of `Al` by `Pb^(2+), E^(@)` is positive and therefore the reaction is spontaneous : Note that we have multiplied the `Pb^(2+)//Pb` half-reaction by a factor of `3` and the `Al//Al^(3+)` half reaction by a factor of `2` so that the elecrons will cancel. but we do not multiply the `E^(@)` values by these factors because electrode potential is an intensive property.
`{:(Pb^(2+)(aq.)+2e^(-)hArr Pb(s)" "E^(@)=-0.13V),(Cu(s)hArr Cu^(+)(aq.)+2e^(-)" "E^(@)=-0.34V),(bar(Pb^(2+)(aq.)+Cu(s)hArrPb(s)+Cu^(+)(aq.)E^(@)=-0.47V)):}`
For the oxidation of `Cu` by `Pb^(2+), E^(@)` is negative (-0.47 V), and therefore the reaction is non-spontaneous.
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    R SHARMA|Exercise Question Bank [Building the Knowledge Level I]|1 Videos

  • ELECTROCHEMISTRY

    R SHARMA|Exercise Question Bank [Building the Knowledge Level 1]|2 Videos

  • COORDINATION COMPOUNDS

    R SHARMA|Exercise Archives|56 Videos

  • GENERAL PRINCIPALS AND ISOLATION OF ELEMENTS

    R SHARMA|Exercise Arechives|16 Videos

Similar Questions

Explore conceptually related problems

Predict whether Ag^(+) can oxidize Pb to Pb^(2+) under standard state conditions. E_(Ag)^(@)=0.799V and E_(Pb)^(@)=-0.126V .

p_(4)(s)+3OH^(-)(aq) . For this reaction the oxidizing and reducing agents are, respectively :

A1^(3+) + 3e^(-) to Al(s) , E^(@) = -1.66V Cu^(2+)+ 2e^(-) rightarrow CU(s) , E^(@)=+0.34V What voltage is produced under standard conditions by combining the half reactions with these standard electrode potentials?

Al^(3+)(aq) + 3e^(-) rightarrow Al(s) E^(@) = -1.66V Cu^(2+)(aq) + 2e^(-) rightarrow Cu(s) E^(@) = +0.34V What voltage is produced under standard conditions by combining the half-reactions witih these standard Electrode Potential?

Determining the realtive strengths of oxidizing and reducing agents a. Arrange the following oxidizing agents in order of increasing strength unde standard-state conditions: Br_(2)(1), Fe^(3+)(aq.), Cr_(2)O_(7)^(2-)(aq.) b. Arrange the following reducine agents in order of increasing strength under standard state condition: Al(a), Na(s), Zn(s) Strategy: Pick our the half reactions in Table 3.1 that involve the given oxidizing or reducing agents and list them, along with their E^(@) value increases (i.e. becomes more positive) whereas the strength of a reducing agent increases as the E^(@) value decreases (i.e. becomes more negative).

STATEMENT-1: In the reaction Zn(s) + Cu^(2+)(aq) to Zn^(2+)(aq) + Cu(s) . Cu^(2+) ions act as oxidising agent and Zn atoms act as a reducing agent. STATEMENT-2: Every redox reaction cannot be splitted into two reactions one being oxidation and the other being reductioin. STATEMENT-3: The oxidation numbers are artifical and are useful as a book keeping device of electrons in reactions.

Assertion : Zn(s)+Cu^(2+)(aq)rarrZn^(2+)(aq)+Cu(s) can be split into following two half reactions underset(("oxidation half reaction"))(Zn(s)rarrZn^(2+)+2e^(-)) Reason : Every redox reaction can be split into two reactions, one representing loss of electrons and the other representing gain of electrons. Cu^(2+)(aq)+2e^(-)rarrunderset(("reduction half reaction"))(Cu(s))