Question
A voltaic cell consists of a standard hydrogen electrode in one half-cell and a Cu/Cu2+ half-cell. Calculate [Cu2+] when E cell is 0.070 V.
NOTE::::::::::::::::::::ANSWER IN SCTIENTIFIC NOTATION PLEASE!!! thank you!
NOTE::::::::::::::::::::ANSWER IN SCTIENTIFIC NOTATION PLEASE!!! thank you!
Answers
Cu ==> Cu^2+ + 2e Eo ox = ?
2H^+ + 2e ==> H2 Eo red = ?
-------------------
Cu + 2H^+ ==> Cu^2+ + H2 Eocell = ?
Then Ecell = Eocell - (0.05916/n)*logQ
where Q = (pH2)(Cu^2+)/(Cu)(H^+)^2
You know Ecell is 0.070v.
You know pH2 = 1 (it's standard H2 electrode).
You know (Cu) = 1 by definition.
You know Eocell from the above calculation which is the sum of the Eo ox and Eo red values.
Post your work if you get stuck.
Solve for (Cu^2+).
2H^+ + 2e ==> H2 Eo red = ?
-------------------
Cu + 2H^+ ==> Cu^2+ + H2 Eocell = ?
Then Ecell = Eocell - (0.05916/n)*logQ
where Q = (pH2)(Cu^2+)/(Cu)(H^+)^2
You know Ecell is 0.070v.
You know pH2 = 1 (it's standard H2 electrode).
You know (Cu) = 1 by definition.
You know Eocell from the above calculation which is the sum of the Eo ox and Eo red values.
Post your work if you get stuck.
Solve for (Cu^2+).
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