Electrode Potential
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Q. Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox Titration. Some half – cell reactions and their standard potential are given below:
MnO−4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(I); E∘=1.51 V
Cr2 O2−7 (aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2O(I); E∘=1.38V
Fe3+(aq)+e−→Fe2+(aq); E∘=0.77V
Cl2(g)+2e−→2Cl−(aq); E∘=1.40V
Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO3)2.
MnO−4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(I); E∘=1.51 V
Cr2 O2−7 (aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2O(I); E∘=1.38V
Fe3+(aq)+e−→Fe2+(aq); E∘=0.77V
Cl2(g)+2e−→2Cl−(aq); E∘=1.40V
Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO3)2.
Q. While Fe3+ is stable, Mn3+ is not stabe in acid solution because:
(Mn3++e−→Mn2+, E∘=1.5 V;O2+4H++4e−→2H2O, E∘=+1.23V;Fe3++e−→Fe2+, E∘=0.77V)
(Mn3++e−→Mn2+, E∘=1.5 V;O2+4H++4e−→2H2O, E∘=+1.23V;Fe3++e−→Fe2+, E∘=0.77V)
Q. Given below are the half-cell reaction:
Mn2++2e⊖→Mn; E⊖=−1.18V
2(Mn3++e⊖→Mn2+; E⊖=+1.51V
The E⊖ for 3Mn2+→Mn+2Mn3+ will be:
(IIT-JEE 2014)
Mn2++2e⊖→Mn; E⊖=−1.18V
2(Mn3++e⊖→Mn2+; E⊖=+1.51V
The E⊖ for 3Mn2+→Mn+2Mn3+ will be:
(IIT-JEE 2014)
- (a) -0.33 V; the reaction will not occur
- (b) -0.33 V; the reaction will occur
- (c) -2.69 V; the reaction will not occur
- (d) -2.69 V; the reaction will occur
Q. The standard reduction potential at 25oC of Li+/Li, Ba2+/Ba, Na+/NaandMg2+/Mg are −3.03, −2.73, −2.71 and −2.37 respectively. Which one of the following is the strongest oxidising agent ?
- Na+
- Li+
- Ba2+
- Mg2+