Electrochemistry Exam Questions and answers 2025.

Electrochemistry Exam Questions and
answers 2025




In .which .of .these .compounds .does .phosphorous .have .an .oxidation .number .of
.+5?




A. .PH3


B. .CoPO4


C. .Mn3(PO4)2


D. .PCl3


E. .P2O5 .- .CORRECT .ANSWER-B, .C, .E


Why .B .is .Correct-
First, .remember .that .polyatomic .ions .in .the .compound .must .have .oxidation
.numbers .that .equal .the .overall .charge. .PO4 .has .a .3- .negative .charge .so .the
.oxidation .numbers .of .phosphorus .and .oxygen .must .equal .negative .3. .Using
.known .rules, .Oxygen .has .an .oxidation .number .of .2-. .Multiply .-2 .x .4 .because
.there .are .4 .oxygen's .in .the .ion .to .get .a .total .charge .of .-8. .Since .P .+ .O4 .must
.equal .-3, .phosphorus .must .have .an .oxidation .number .of .=+5.




Co= .+3
P= .+5
O= .-2

,Why .C .is .correct-
It .is .important .to .recognize .the .subscript .changes .the .amount .of .elements .in
.the .compound. .Using .the .known .rules, .we .know .that .O .is .-2 .and .Mn, .since .it .is
.in .group .2 .would .be .+2. .Multiplying .through .with .the .subscript. .we .get:
Mn= .2 .x .3= .+6 .and .
O= .-2 .x .4 .x .2= .-16


To .calculate .the .oxidation .number .of .Phosphorus, .remember .that .there .are .2 .so
.that .should .be .accounted .for.
+6 .+ .2x .+ .-16= .0


x .must .be .equal .to .+5


Why .E .is .correct- .
O .= .-2 .x .5= .-10
P= .2x= .10 .to .equal .0. .X= .-5


The .oxidation .number .of .Mn .in .Mn3(PO4)2 .is .- .CORRECT .ANSWER-+2


It .is .important .to .recognize .the .subscript .changes .the .amount .of .elements .in
.the .compound. .Using .the .known .rules, .we .know .that .O .is .-2 .and .Mn, .since .it .is
.in .group .2 .would .be .+2. .
Mn= .2 .x .3= .+6 .and .
O= .-2 .x .4 .x .2= .-16
P= .2 .x .5= .10


The .oxidation .number .of .hydrogen .in .AlH3 .is .- .CORRECT .ANSWER--1


AlH3
H= .-1 .x .3= .-3
Al= .+3


In .Sn .(s) .+ .2HCl .(aq) .⟶ .SnCl2 .(aq) .+ .H2 .(g), .what .is .being .oxidized .and .what .is
.being .reduced? .- .CORRECT .ANSWER-Sn .is .being .oxidized, .H2 .is .being
.reduced.

, 1. .Assign .oxidation .numbers .to .all .atoms .in .the .equation .and .ignore .the
.coefficients .in .the .equation. .
.Sn .(s) .+ .2HCl .(aq) .⟶ .SnCl2 .(aq) .+ .H2 .(g)
.0 .+1 .-1 .+2 .-1 .0 .




2. .Compare .oxidation .numbers .from .the .reactant .side .to .the .product .side .of .the
.equation. .If .a .redox .reaction .has .occurred, .you .will .find .that .the .oxidation
.numbers .of .two .(no .more/no .less) .elements .have .changed .from .the .reactant
.side .to .the .product .side. .The .element .oxidized .is .the .one .whose .oxidation
.number .increased. .The .element .reduced .is .the .one .whose .oxidation .number
.decreased.




Sn= .0 .⟶ .Sn .= .+2, .tin .is .oxidized
H= .+1 .⟶ .H= .0=, .hydrogen .is .reduced


Is .ZnCO3 .(s) .⟶ .ZnO .(s) .+ .CO2 .(g) .a .redox .reacrtion? .- .CORRECT .ANSWER-No,
.nothing .is .being .oxidized .or .reduced.




2 .Ga .(l) .+ .3 .Br2 .(l) .⟶ .2 .GaBr3 .(s)
In .this .reaction, .what .is .the .oxidizing .agent .and .what .is .the .reducing .agent? .-
.CORRECT .ANSWER-Br2 .is .the .oxidizing .agent
Ga .is .the .reducing .agent


1. .Assign .oxidation .numbers .to .each .compound .in .the .reaction.
2 .Ga .(l) .+ .3 .Br2 .(l) .⟶ .2 .GaBr3 .(s)
.0 .0 .+3 .-1




2. .Compare .oxidation .numbers .from .reactant .to .products.


Ga= .0 .⟶ .Ga= .+3, .Gallium .is .oxidized, .therefore .it .is .the .reducing .agent .since
.the .electrons .it .loses .reduced .the .charge .of .Bromine.
Br= .0 .⟶ .Br= .-1, .Bromine .is .reduced, .therefore .it .is .the .oxidizing .agent .since
.the .electrons .it .gains .are .a .result .of .Gallium .losing .electrons.




For .this .reaction, .which .is .the .correct .unbalanced .oxidation .half .reaction?