Calculating cell potential (Eeenl
1. Consider an electrochemical cell (a.k.a. galvanic cell, voltaic cell, battery) with Zn(s) and
0.25 M Zn(N03)2(aq) in one compartment and Cu(s) and 0.25 M Cu(N03h(aq) in the other
compartment.
a. Calculate the cell potential (EMF, voltage) at 298 K.
Reduction (cathode): Cv""(C%"\) -t
f1---:,
(u (S
1
p-
Oxidation (anode): C.III (S) -:IJ (0/1 'l*(Oq, of
Net (overall cell reaction): 2""(51 t C\.)1.""co.o.)~-ZIl)H(oe;)+(u«5)
EcelJ
~celJ- (0.0592/n)10g Q =
q'OI 1J.
-\-
I.IQO V
te~(ll ;: ~ .IClO V1~ 'SroV;t-q(\fOJ)
C
(. 100 V -
EOcell=
V
""-.:.2
( e\..e1.A ~L.. tLl<j I~ ~ t a s+o....dnrd c"tH) ~ ~eol~~
~:,(.odt~ l \~ 0. stq",;-rd C~\ I \o~ co lise Q::. \ ~ )
b. What would be the cell potential if the concentration of Zn(N03)2 was increased to
2.5 M?
(}::
~ ~ ~ = tu
[Cu"l.'l
(J.1...C; Ec.dl = (. l 00 V
-
lC().J (I())
Q.05\12 vi
2.
A ((j-fo\~ 1V'\<-~qS~
G< ~slAts
lll\
f.o 76 V
:=
\11\ a
\.R."') 'S~(l v(Jttq("~~ C~·~. c. Calculate the equilibrium constant (at 298 K) for the overall cell reaction.
6~O:
-
V\
F E: .. \l
;.
-(2)( q6 IS-cO ':,\}(
-=-
-112.3
(2l1.?
E<-elt:::.
a
;1';: - 2
(1/ ,CO
,..~(
0
~c..' t:.s(.w.i)
t6.2,
I "f)(
iG·'3
~
g~c:qU ~tA+
(.IOD
u,)4tJ\
~
:iJe 2~1 k:) ._
-
'f(.OC+ICJ-\ t")
tl{
~i
I.G'1><IO
I(
,
.
~\..Jl LlorwifV'( 6)-::.. ~ ')
pl\J~ ~\'S \)o.l~ c~ ~~t:\ \""+0 t~ N~jN7+ ~<l['-lQt-,oV) ~.",d
~t.\)
c
~"'(,u{6 ¥t- E'~tI=: d.
~u~ cJ,,<:t~
Lv..
Now
yt/\J
Co.VI
s~~ w"'~ \t to.~.Q.5
~ tc::. ~C-"""~ E~tI ~\1lAl ~,co.J.I~,
6
2. Consider a voltaic cell with Cr(s) and Cr3+(aq) in one compartment and Zn(s) and Zn2\aq) in
the other compartment.
a. Draw and label a diagram of this cell.
LIA
t'5
t l~"'l C.,-)
VV'c~ ct<.hl..R.
cr...
'2."", (~ t~ QV\od~.
J-'
'--------~.013 \lJ~
N"," "---:p
2~
ZV\(')~-:It -Z.., 41-e(
T~
\le:,\
cr3+-(Qct)4,t-~ er(')')
tG,<J~ o.S'S\J~" -tL..c..-+ ~ ':. \ )
b. Calculate the standard cell potential and the equilibrium constant for the overall reaction.
re.d :
1)(
(c("" '3+(Q~) + ., e-
0)(, '. ?i~ ( """ZV\
It\~t: :'''2.11'\('))
Co:; ')
"2.+
~ "ZVl
-O.'1t/
(0<'".,1 t 1~)
Ec.o",od =-
- 0.1
2-t £~
Cr ('))
- \
;4
+ 2C'(" (O-q)~ ?,L""
tl:/' - V\ t E:..\ ~
::
_ l:/s'C./'f."T..
1< ~C\ -:: ~
( NCJ-t~:
£:.."l.::'
--1'
;:
o..t Lct~ k."}
E~", t~()'-1 t, -t~~
sttt{
~ ~~c..+toV\
(o~l+ 2crC-:';'
e~1l ,"2") V
;:. +a 02) V
-(6') (qbISOO~ c.O"2.,
Ie
V
\
.
l )
t..-.,. 5 rVl+Q",e(jU~ (100.1":. L-)
i ~ ;:. - I"h,n ~I
~r
- t3.:' J.. ;;l
e.
12- l'
1~."'1 ~ ji,'{'b ~:\ ~ - 2 t s­
::. 1.~~ \::1'
/'"'.,.\ \
dl~~re-<~ lVl t~ ~d~C;~I<',""" Io<.t~",,·h::Js
I
~~) ~ f~\t) \?t~ r:~9 ~ t
lj 0""
,..
J '"'
<:
"",co II
..J
7
c. Calculate the cell potential if [Zn 2+] 5.00 M and [CrJ+] = 0.50 M? What does this result tell
you? How would your diagram change? How does Q compare to K?
Q = [ "t~"'''13
_
[C<)~J2
~
C
::S'.Ov1 ;:
(0:50')2
£C... II;;. Ee(.~!t . - ,o.o!.~V
.....
Not S?GIII'\-C\oJ\eouS
t--bw
Q -;> KeG
( SCO "";) L l5
d. If zinc is the anode and [Cr3+]
0.0$0 V
~.::
:=
G.O"1..JV ­
_")
1.~:,y.X'(()
.
[2V1~1 :::
Sc dln).,.-o."""'" ~ ~ rSe. .
C'\('If"ClW'5 l ""
/\ Q
- Q.OO,G ~ V -
,A,ll
(oq
J
V\ = b
{trG
500
J
0.0 tOM, what must the [Zn2+] be in order to achieve 0.050V?
0. 0 5£12_"
b
t0') Q
C
LVI"1."'J;'
;~
(C.C>\CI'\
(~I
'S. '15)<\0 !'
w4.... Q~
£(.~If::
I ~
t
a.O,O v
0.((
C"""
~ut- a\/l~
cOv-C.Q"'->+r::'-,,+tCn,
{oil., If>u tl.,
«.V1GW
V1
)
tL~ V\
u~t.~V' C",x--"...,kt'G~.!
(5~ i~ ~~ ~rCJlJl~y\.
J
e. What must the value of Q be in order for the voltage to be zero?
Q::
~e~::.
2\S
8
3. A standard zinc half-cell was coupled to a standard hydrogen electrode. When the red wire was attached to the hydrogen electrode, the cell potential was 0.763 V. What is the standard reduction potential for the zinc half-cell? [Zn2+] = 1.0 M, [H+] = 1.0 M, P H2
Reduction (cathode):
Oxidation (anode):
Net (overall cell reaction):
J H~ ~
1 atm (standard hydrogen electrode)
.
2e- --==t ~:l 7.1/\ ~ "ZV1 '1>+ ~ :2.e-
LV\ + -Z 1-1 +~ L.fl '2.+ -+
O.dO V
ganode
X
Wz
Tl.\'S IS Ir,c--.J
G~
gcathode =
VV'Q<::I.S0
~
,:.,b..,cbrcl
~duc:4")GY\ rck~+'c: 1<;
vr::c) {
4. A Zn/Zn2+ half-cell was coupled to a hydrogen electrode in which Pm 1 atm. The [Zn2+] in
the anode compartment was 0.10 M and the cell emf was 0.542 V. Calculate the pH in the
cathode compartment.
[-2.V11.t")::=.
CL! 0 f0
f H2­ -:= l ,OOIt~
Cc"~ +\...c/:h:~
(-l
(-1-\
~-_~'~h"""-'- tI::.. "'viol,. ( <:><:,
G - \ - - p.,..
2 Nt- -1 2 ~- .....,.
~
E~lt
t12 :;: O.5tj'l.
V
~d,
( \t-,,,
I.,
",,,+ Q<-
G
r Q f.
-r+ ,.., '"
/1"",<<0"
Q
9
~)