Table of Contents
Common Decrease Capacities
The adhering to< img align=" absmiddle" boundary =" 0" elevation 
=” 57″ src =” https://mmsphyschem.com/wp-content/uploads/2018/04/estandardtag-1884283.gif” size
=
” 161″ > might serve. 1) Calculate E ° for the adhering to response.
2HNO3( aq) + 6HBr( aq)
< img elevation=" 12" src=" https://mmsphyschem.com/wp-content/uploads/2018/04/arrow-6902772.gif" size=" 43" > 3Br 2( l)+ 2NO( g) +
4H 2 O( l) (
a) -0.106 V
( b) 2.026 V
( c) -2.206 V
(
d ) 0.106 V( e ) -1.278 V 2) Calculate E ° for the adhering to response
. Sn( s) + 6F -( aq) +4Fe 3 + (aq)< img elevation=" 12" src=" https://mmsphyschem.com/wp-content/uploads/2018/04/arrow-6902772.gif" size=" 43" > 4Fe 2+ (
aq) +[ SnF 6] 2 –
( aq )( a) 0.521 V
( b) -0.521 V
( c) 1.021 V
(
d ) -1.021 V( e ) 3.334 V 3) Calculate E ° for the adhering to response
. 2Tl+( aq) + 2I–( aq) < img elevation=" 12" src=" https://mmsphyschem.com/wp-content/uploads/2018/04/arrow-6902772.gif "size =
” 43 “>
I 2( s )+ 2Tl( s) (
a) -0.875 V
( b) -1.215 V
(
c) 0.195 V( d) -0.195 V ( e )0.875 V 4) Calculate E ° for
the complying with response. TWO( s) +4H +( aq)
+ Zr( s )< img elevation=
” 12 “src =
” https://mmsphyschem.com/wp-content/uploads/2018/04/arrow-6902772.gif
” size=” 43 “
> Zr 4
+ ( aq )+ 2H 2 S( aq ) ( a) 1.81 V (b) 1.67 V( c) -1.39 V( d)
1.39 V( e) -1.67 V. 5) CalculateE ° for the adhering to response
. 2Al 3
+( aq )+ 3V( s )
< img elevation=
” 12 “src =
” https://mmsphyschem.com/wp-content/uploads/2018/04/arrow-6902772.gif ” size=” 43″ >3V 2+( aq) +2Al( s)( a) -0.48 V( b) 0.22 V( c)
-2.84 V( d) 2.84 V( e) 0.48 V 6)
+ 6H +( aq )
+ 2Au( s )< img elevation=
” 12 “src =
” https://mmsphyschem.com/wp-content/uploads/2018/04/arrow-6902772.gif” size=” 43″ >
3H 2
Se( aq )+ 2Au 3+( aq)( a ) 1.90 V( b )-1.10 V( c) 1.10 V( d)
-1.90 V( e) -4.20 V 7)Calculate E ° for the adhering to response. 6CO 2( g)+ 6H+( aq)+ 2Al( s)< img elevation=" 12" src=" https://mmsphyschem.com/wp-content/uploads/2018/04/arrow-6902772.gif" size="43” > 3( COOH)
2( aq )+ 2Al 3+( aq)( a) -1.17 V(
b) 2.15 V
( c) -2.15 V
( d) 1.17 V
( e) 1.85 V
.
8) Calculate E ° for the adhering to response.
Te( s) + 2H2O( l) + 2Sn4+( aq)
+ 4H +( aq) +
2Sn 2+( aq)( a) 0.229 V( b
) 0.379 V
( c) -0.379 V
(
d ) 0.679 V( e ) -0.229 V 9) Calculate E ° for the adhering to response
. Na 2SO4( aq) + Pb( s)
” 43 “>
PbSO 4( s )+ 2Na( s)( a
) 5.072 V
( b) 2.358 V
( c) -2.358 V( d) 3.07 V(
e ) -5.072 V. 10) Calculate E ° for the complying with response.
NiO2( s) + 2Fe2+( aq) + 4H+( aq)
+ H 2 O (
l) (a) -0.929 V
( b) 2.471 V
( c) 0.929 V
( d) 0.158 V
(e) -2.471 V.
