00:01
Hi, hello, see the answer for given question.
00:03
So first we need to write anode half cell reaction at here.
00:06
So first of all, so a part i am doing.
00:09
So for this, we need to write anode half cell reaction.
00:15
So anode of cell reaction is becomes what? anode of cell is reaction, nothing but here oxidation of copper.
00:25
Okay, so anode half cell reaction is oxidation of copper.
00:30
So for this oxidation of copper i am writing the equation so cu gives rise to cu plus 2 plus two electrons minus the e not cell value for this is a minus 0 .34 volts.
00:48
Okay, so after that i am writing that cathode of cell reaction so cathode half cell reaction so the cathode half cell reaction so the cathode half cell reaction so the cathode half cell reaction for this will become what the reduction of silver okay reduction of silver it will be represented by ag argentinum so that one i am writing to a z plus okay plus two electrons and gives rise to two a z the e not cell value for this one is okay so plus 0 .80 old so plus 0 .80 old so now i am writing the overall reaction okay so overall reaction i am writing so the overall reaction means by combination of both oxidation and reduction half cell reaction i am writing complete cell reaction so cu plus 2 electrons and for silver to a z plus plus two electrons gives rise to 2 a z so the e0 value for this one is minus 0 .3fold volts.
02:08
The e not value for this one is so 0 .80 volts...