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@Dexter
Note for OP
+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.yeahhh so first we find the no of moles of e = 5.79 x 10k / 96500 which gives 0.6 moles
and then at cathode the cu2+ + 2e- --> cu and the moles of e is 0.4 moles
i didnt get what we do at anode
it gives o2 gas yeah
uske baad samaj nahi aayi
yeah but the doubt i had is that, how are we considering 0.2 moles of e- to be 0.1 mole of h2 at the cathode
and why are we considering the no. of moles of e- calculated initially for calculating the o2 gas moles and we use the molar solution conc in calculating the h2 gas moles
why cant we use the same amount of moles in both cathode and anode
like the same value
ohh okay okay got it 👍
haan equivalence mera backlog hai 💀 l
Check out my sessions on mole concept. Part 5 and part 6.
okay okayyy
+solved @iTeachChem
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