I<E~-

5.3 Notes: Electrochemical Cells

-OOKU-P hO\"t7Ce\\ fxnjI"n --\ab\~ '.

H-' 9\te'f nO\ \+ - ce\ \ r xni.. ~e. (a\-'r\\)de.,··

Z.r '2--1- + '2.e- ~ Zn (s)

\ AnoQ.0 1-

2()(s) ~ 2('?~t-t 'L«

Definitions

Zn ----I>l:;ro

1 MZn(N03)z2n1-t -t'LtJO;

Cu

1 MCu(N03hCu '2. -\ -t 2 NO~

A- (1 \\ nQ~j _ Electrochemical cell- A device which converts chemical energy into electrical energy2 e' tC\,Cl {}J.:S ' Electrode - A conductor (usually a metal) at which a half-cell reaction.

i (oxidation or reduction) occursAno le.. O,na Anode - The electrode at which oxidation occurs. (A & 0 are both vowels) LEOA(label the anode in the diagram above)

Gc~hod~, Cathode - The electrode at which reduction occurs. (R & C are both consonants) GERC• (Label the anode in the diagram above)

Half-cell reactions C ~ +h 0..t I'~ L..0WE- ~ \ II i-a bi e) .Meta\ .s +\t)~ w(\9.er gA

fe~e\'Se. (oi;clahcn) ~ -l-nL~) -7 is(I'7--1 + 'U--Anode - Oxidation half-rx:

Metal atoms are changed to + ions. Metal dissolves and anode loses mass as the cell operates.

Cathode _ Reduction half-rx: 2-n1-t ion S OLA" ~ u, P \f\ b e0 \<..e r .

+ ions are changed to metal atoms. New metal is formed so the cathode gains mass.

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Flow of electrons-Since the anode loses e':s, (LEOA) and the cathode gains es (GERC)

Electrons flow from the anode toward the cathode in the wire (conducting solid)

LEO--Cathode - Cu~+G~CU

:..==:::::::~ L..=-=-=-=-..l__ ~ , t I'C ~ "\O&~ J p\1l~m:\.1e-'5 flow from H C ;n the w;re e ~ 130;~ ICOil ~wnWFlow of Ions in the salt bridge

-Salt bridge contains any electrolyte (conducting solution: ionic solution or acid or base)

Porous barrierhich ions can

pass through

Common things which contain[electrolytes, e.q.) potatoes,japples, oranges, lemons,wrogs, people

Most common - U-tubecontaining a salt such asKN03or NaN.P3'.'~4 8)\\.\b\e &l\t

~ \\.-.\.-\io(\

As cations are produced atthe anode, they build up andcreate an excess of positivecharge.

An/Dns flo~ fo ~ODe--(0.+\ o~s .f low to C~pro(JE----

Ecailiode JCu2+ + Ze' -7 Cu

As cations ions are depleted at thecathode, there becomes moreanions than cations. An excess ofnegative charge builds up.

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Cations (+) flow towards the cathode }Anions (-) flow towards the anode in the salt bridge

Show with arrowsthe , , ,~kection of flow of all ions '

2+ NO - K'+ C 2+ '(Zn, 3" U »iAlscishow thedirectionof 'iflowof electrons in the wire

CuZn'1-\ .-

mCS) ~ 1:.0 +le

Identifying the ANODE and CATHODE- Look at the reduction table- All half-rx's are reversible (can go forward or backward)- The half-rx with the greater potential to be reduced is higher on the table

(higher reduction potential EO)

***So the HIGHER half-rx is the CATHODE (reduction) ,LOWB~ "a\~ -'(1- ,S A~OD6 (~h:oll~e'(RA ... m,v(1{ \i'N,\L,\

Ex: Cui Zn to '\:::It O~ID\~ED) ~Cu2+ + 2e- -7 Cu higher (gets to be cathode).72++k~zb

The 4NODE rx wi1l2~ere~ersed (oxidation): , ,\(Anode rxrZn rs Zn +2e) LEO l \()<);~ da~)

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~"1../ CA -\--\\0 at, "

~--tT::}fl

Important Note:The # of es involved in the oxidation (ANODE) must = # es involved in reduction(CATHODE) }Jl"1.4 ~2e- -? Ni(S) is h<g\v( ,"f' tab\G- ' .Ex: Cathode (Red):3 (N\ .2..-4 + 1e-c

- ~ N; (s))Anode (Ox): LeA \ (<, ) ---7 A \ 1+-+ :3e.1

A}.IoDt" Net Ionic Equation for REDOX rx:

'3N~).-\-r~ ~ 3'Nl(~)+ 2... A \ C~) ? 2. A \ '3 -t -\-y(

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Hebden Q's pI 217 # 34 (a-e), 35 (a-e)

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Standard reduction potentials and voltagesVoltage - The tendency for e-'s to flow in an electrochemical cell.

Basically the work done per electron transferred (from half-cell to half-cell)ANOD£. ~ CI4TtfOO£'

Note: a cell may have a high voltage even ifno e-'s are flowing. It is the tendency(or potential) for e-'s to flow.

In practice, we measure the difference in electrical potentials between 2 half-cells.EOc-e\\

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Reduction potential of half-cells-The tendency of a half-cell to be reduced. (gain e-'s)

Voltage only depends on the difference in potentials not the absolute potentials.Thevoltage{Qtacgll:depepdsorilyonth~difference)inreauctiorfllotel'1tiais afthe' tWohalf~Cells.A,'1%{''',·'f

','" . , .... v.··.:,',··",.' ,.i::,:'" .. " .. ,,' .',. '"

. 0r;; ({'II ::: E b ~e.D- E 0Dl'.

Relative potentials of half-cells can only be determined by connecting with otherhalf-cells and reading the voltage.

E.g.) How good a basketball team is can only be determined by playing withother teams and looking at points (scores).

The Standard Reduction Potential (EO)- A "standard" half-cell was arbitrarily chosen to compare other half-cells

with.- It was assigned a "reduction potential" of 0.000 v

- It is: 2H\aq) + 2e- E ;, H2(g) EO = 0.000 v

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What is standard state:- 25°C ,- 101.3 kPa (1 atm)- 1 M concentration for all solutions involved in half-cell

The Standard Hydrogen Half-cellH2~ 2e-+2Ir

IConnecting wire

Bubbles of H2

H2 (g) (at standardpressure 101.3 kPa)

1.0 M H+ at 25°C

Platinum electrode

INERT electrode: provides asurface for reaction to occur.

Pl, c.P-L -e\~~cl~ (Un ClC'(~pt.or dODa-J(;

e-~ o: ~ f\-( -e.d..-Ld .

Looking at the Reduction Table ....All the voltages (reduction potentials) given in your table have beenfound by connected another half-cell to a hydrogen half cell. The voltagewas measured, relative to the hydrogen half -cell.

Ex: The half-cell made by placing Cu metal in a solution of 1M Cu2+, thenconnecting this half-cell to the hydrogen half-cell had the voltage:

EOCe\\'-=--tO,~~ C 2+ + 2 - ~ C EO= +0.34 V--E!J u ,e =r uCs) ~~

-tY\,s. 'j S -\-~~ \fa \~-( ~O\.\

~-e e. -\-cr (\.."".'2-\ 4- 1e -=7CU 0 )

o f\ -\-C\'0 \e. :.J

.-\- () I~L.\ '-l (Y\ Of c. ~ \rt 0 "

t-\ "}... 'v\CA l\ (e \ \ . 6

Reversing reduction reactions: tNote that all reactions in the table are written as reductions (forward)

If we reverse the reaction, we have the oxidation reaction:

Red: Zn2++ 2e- -7 Zn(s)

'2 2 1..t 2-Oxid: (\ (~") -7' (\ A (,

If a half-rx is reversed, the sign of EO is reversed too.

r ~IV\e-\- \ O(\\CEx #1: Calculate the potential of the cell (EOcell): Ni2+ + Fe -7 Ni + Fe2+e o ~

\ t \)" ~A-

The half-reactions involved (as they appear in table):

Ca\V\oat t~~(,\)Ni2++ e- -7 Ni s) EO= - 0.26 V

A"""ode, (Red') Fe2+ + 2e- -7 Fe(s) rEO=- 045 V~ .II __ ----._-- fe\Je~ ~. ~(-eve,<..z~ljn

__ ..:::;, Method 1: Write the Fe2+/Fehalf-reaction as oxidation (reverse)

Cu~h {fRd} , Ni2++2e- ~Ni(s) EO=-0.26V \\ADD 1/

An ~ (~ J~!e(S) ~ Fe2+ + 2e- t EO= + {J \ 4-5 \j-V- -'

r~\j-l vUd LEO reIJ ~ -\- O. \ q" .

Joining Half-Cell Reactions (Making an Electrochemical Cell!)D\ r-F b \vJ E 0 ''J Ot t\.( ('s

We calculate EOcellby finding the 'voltage gap' between the 2 half-reactions.o~\cA,·

c~=:-7 Method 2: EOred- EOox = EOcell (write exactly how they appear on table)

CAt~ol}cltA(\,: tJ.{ FI'(\d ~lrle c\ it\-~r.i (\ <.e _F-~{" -::::-E(fo\ - Eo~ bfLtWPln (l'dUCh'Of)

. "PO.J..e n ~lal S .I ~~~\i =. T-"/Q), 2.b \l - (- D.4-5\J) ('Bo~n EO o.\U{~os V.eclU(;h")~)

fd- ~ "~ -t O. \ q" \ .\C t·c:\ \

o 'f.-I c\} a (\OC\ 0{ -..\1

Ex #2: Calculate the potential of the cell (EOcell): 3 Ag" + Al 73Ag + AI3+\ 1-

re c\ :>l(1-\l(\tO..Q.,

The half-reactions involved (as they appear in table):

eG\\hO~t ~ A~+-te- ~ A9(\) EO=O,'tloV

A(\(Jc\ e '. A \ ~-\: -+ 3e- ~ A \ (S ) ~ 0 ~ - \ ••bt, \Note: to make eo's lost = eo'sgained, we multiply Ag+/Ag half-cell by 3.

BUT... E 0 do es n()~ Change (dorl'{. ..~ ~) .Method 1: Write AI/AI'+as oxidation reverse) and add: J

CCl\Y\Od.~ 3A3 -\- ~) -'73Ag uj E"-.~

A\('~) ~ Al~-t+~~. EC-::EfJI-Ib\l

E"C~l\ =: b, ~..."L-\- b \ITEU ClII = O:~O \I - (- \ .bb \] ) =- C-+ 2-.L.\-b'-l J

AY\c(i~('{.~\je\~t~·D~')

Method 2:

Spontaneity

The 'sign' of the cell potential tells us whether the redox reaction will bespontaneous:

EOcell = +Eo -

cell- -

SPONTANEOUS C ceu do·es w od~)NON-SPONTANEOUS (\'\JOY'k mu..st bCL

c\Ol{\t t-o ~~SA·Cn1 tojet (.e\\ 1D opuakJ·

Rehden Q's p. 224/225 #36, 37

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