Question

Answer third question....

Answer 1

$\huge \: \red \star \: { \large \pink { \underline\mathcal{ANSWER}} } \: \red\star \: </p><p>$

reduction involve loss of electrons and oxidation involve gain of electron

trick to learn :- OIL RIG( short form)

full form = Oxidation Involve Loss

and Reduction involve gain

at anode electode oxidation takes place and at cathode reduction takes place

trick to learn :- LOAN( short form)

full form :- left oxidation anode negative

since given half cell potential are standered potential or standered reduction potential

$\\E^{\circ} _{Al|Al^{+3}} \: = \: - 1.66 \\ E^{\circ} _{Sn|Sn^{+2}} = 0.14$

Sn is upper then Al in electrochemical series [ in which flourine is on top and li is on bottom ]Sn strong oxidising agent then Al

so Sn oxideises Al and reduce themself

and at anode oxidation takes place so

the anode electride is made up of

$\\ \boxed{ \green{Al}} \\ \:$

also

emf or cell potential is

$\\ E^{\circ} _{cell } \: = \: E^{\circ} _{cathode} - E^{\circ} _ {anode}$

so emf is

$\\ E^{\circ} _{cell } = 0.14- ( - 1.66) \\ \: \: \: \: \: \: \: \: \: \: = \boxed{ \green{ 1.80 \: v}}$

$\boxed{ \boxed {\boxed { \red {Thanks \: \: \: dedo \: \: \: please }}}}$