ENDGAME Q.10

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ramiro_romero
Posts: 90
Joined: Sat Sep 07, 2019 12:16 am

ENDGAME Q.10

Postby ramiro_romero » Sat Mar 14, 2020 7:03 pm

I do not unerstand why the cell potential for 10 is 0.43 volts when the oxidation reaction of silver is reduced. I thought when we reversed an oxidation reaction we changed he sign of its cell potential. Therefore, I thought the answer was (1.23V)-(-0.80V)= 2.03V

Jiyoon_Hwang_2I
Posts: 101
Joined: Sat Sep 14, 2019 12:17 am

Re: ENDGAME Q.10

Postby Jiyoon_Hwang_2I » Sat Mar 14, 2020 7:07 pm

I think when you use the equation E cell = E cathode - E anode you don't use the flipped cell potentials. Instead, you use the cell potentials that were given in the table so it would be E = 1.23 - 0.80 = 0.43. Otherwise you can add 1.23 + (-0.80) which is also 0.43

Indy Bui 1l
Posts: 99
Joined: Sat Sep 07, 2019 12:19 am

Re: ENDGAME Q.10

Postby Indy Bui 1l » Sat Mar 14, 2020 7:12 pm

You don't have to flip the values in the table. You can leave them and just plug them straight into the Cathode-Anode Equation.

Jessica Chen 2C
Posts: 103
Joined: Thu Jul 11, 2019 12:17 am

Re: ENDGAME Q.10

Postby Jessica Chen 2C » Sat Mar 14, 2020 10:42 pm

If you flip the sign then you add the cell potentials; if you’re subtracting then just use the given reduction potentials without flipping.


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