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Question
A copper electrode is dipped in 0.1 M copper sulphate solution at 25°C. Calculate the electrode potential of copper.
[Given: \[\ce{E^0_{{Cu^{2+}|Cu}}}\] = 0.34 V]
Solution
Given that [Cu2+] = 0.1 M
\[\ce{E^0_{{Cu^{2+}|Cu}}}\] = 0.34 V
Ecell = ?
Cell reaction is
\[\ce{Cu^{2+}_{( aq)} + 2e^- -> Cu_{(s)}}\]
Ecell = `"E"^0 - 0.0591/"n" log (["Cu"])/(["Cu"^(2+)])`
= `0.34 - 0.0591/2 log 1/0.1`
= 0.34 – 0.0296
= 0.31 V
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