Test #2 question 4

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lauren chung 2f
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

Test #2 question 4

Postby lauren chung 2f » Fri Mar 16, 2018 12:28 am

On Test 2, it asks to calculate the standard potential for the galvanic cell,
U (s)| U+3 (aq) || V+2 (aq) | V (s)

Since
U+3 + 3e- --> U E= -1.79
V+2 + 2e- --> V E = -1.18
but U is the anode, so I made the equation
U --> U+3 + 3e- E= +1.79

Then I wrote
Ecell = Ecathode - Eanode
= -1.18 - (1.79)
= -2.97 V

Can someone tell me where I went wrong?

Xin He 2L
Posts: 38
Joined: Fri Sep 29, 2017 7:05 am

Re: Test #2 question 4

Postby Xin He 2L » Fri Mar 16, 2018 12:32 am

You don't flip the signs of the cell potential when you flip the equation.

lauren chung 2f
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

Re: Test #2 question 4

Postby lauren chung 2f » Fri Mar 16, 2018 12:51 am

Can you explain why you don't flip the signs of the cell potential?

Rachel Wang
Posts: 49
Joined: Fri Sep 29, 2017 7:04 am

Re: Test #2 question 4

Postby Rachel Wang » Fri Mar 16, 2018 4:57 pm

For the equation Ecell = Ecathode - Eanode, it assumes you have not already flipped the sign of the anode. Therefore Ecell = -1.18-(-1.79).
However, you made the equation E=+1.79. In this case you have already flipped the sign and hence the eq is -1.18+1.79.

Cristina Sarmiento 1E
Posts: 52
Joined: Wed Nov 16, 2016 3:02 am

Re: Test #2 question 4

Postby Cristina Sarmiento 1E » Fri Mar 16, 2018 5:03 pm

When you flip the sign because you flip the equation, you must do Ecell = Ecathode + Eanode.
It is also helpful to know that for galvanic cells, the Ecell must always be positive.


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