Question

The correct representation of Nernst's equation for half-cell reaction \[\ce{Cu^{2+} (aq) + e^- -> Cu^+(aq)}\] is ______.

Options

• `"E"_("Cu"^+,"Cu"^(2+))^circ = "E"_("Cu"^+,"Cu"^(2+))^circ - 0.0592/2 log  (["Cu"^+])/(["Cu"^(2+)])`

• `"E"_("Cu"^+,"Cu"^(2+))^circ = "E"_("Cu"^+,"Cu"^(2+))^circ - 0.0592/1 log  (["Cu"^+])/(["Cu"^(2+)])`

• `"E"_("Cu"^+,"Cu"^(2+))^circ = "E"_("Cu"^+,"Cu"^(2+))^circ - 0.0592/2 log  (["Cu"^+])/(["Cu"^(2+)])`

• `"E"_("Cu"^+,"Cu"^(2+))^circ = "E"_("Cu"^+,"Cu"^(2+))^circ - 0.0592/3 log  (["Cu"^+])/(["Cu"^(2+)])`

Answer 1

The correct representation of Nernst's equation for half-cell reaction \[\ce{Cu^{2+} (aq) + e^- -> Cu^+(aq)}\] is `underline("E"_("Cu"^+,"Cu"^(2+))^circ = "E"_("Cu"^+,"Cu"^(2+))^circ - 0.0592/1 log  (["Cu"^+])/(["Cu"^(2+)]))`.

Explanation:

General Nernst equation for a reaction is given as

`"E"_"cell" = "E"_"cell"^circ - 0.0591/x log  (["P"])/(["R"])`

For the half-cell

\[\ce{Cu^{2+} (aq) + e^- -> Cu^+ (aq)}\]

The correct Nernst's equation is

`"E"_("Cu"^+,"Cu"^(2+))^circ = "E"_("Cu"^+,"Cu"^(2+))^circ - 0.0592/1 log  (["Cu"^+])/(["Cu"^(2+)])`