There are two copper blocks sitting in the $\ce{Cu(NO3)2 (aq)}$ solution, a battery is attached onto both of them, providing enough energy to start the reaction.
Since solid pieces of copper are involved, $\ce{Cu}$ must be considered in the reduction potential as well.
However, looking at the half reaction for copper: \begin{align} \ce{Cu (s) &-> Cu^2+ (aq) + 2e-}& (E^\circ &= \pu{+0.34 V}) \end{align}
Compared to nitrate: \begin{align} \ce{NO3- (aq) + 4H+ (aq) + 3e- &-> NO (g) + 2H2O (l)}& (E^\circ &= \pu{+0.96 V}) \end{align}
Since it produces more energy, I believe the second half reaction will occur instead.
I begin by finding all the species/ions I have:
- $\ce{Cu^2+}$
- $\ce{NO3^-}$
- $\ce{H2O}$
Taking a look at the Standard Reduction Potentials at 25°C:
For my SOA (strongest oxidizing agent) half-reaction ($\ce{NO3^-}$): \begin{align} \ce{NO3^- + 4H+ + 3e- &-> NO + 2H2O}& (E^\circ = \pu{+0.96 V}) \end{align}
As for my SRA (strongest reducing agent) half-reaction ($\ce{H2O}$): \begin{align} \ce{2H2O &-> O2 + 4H+ + 4e-}& (E^\circ &= \pu{-1.23 V}) \end{align}
For the balanced reaction I have: \begin{align} \ce{4NO3- + 4H+ &-> 4NO + 2H2O + 3O2}& (E^\circ &= \pu{-0.27 V}) \end{align}
Some observations that can be made are that:
- Nitric gas is forming at the anode
- Oxygen gas is forming at the cathode
- At least $\pu{0.27 V}$ must be put in
My question is:
- Did I do all of this correctly? There's no answer key and I'm pretty sure that I might have made a mistake somewhere.
- What would happen if you placed a necklace at the cathode?