Write the half reactions and the balanced
equation for the galvanic cell
Cu(s)| Cu2+(aq)|| Cu+(aq)| Cu(s).
What is the smallest possible integer coefficient of Cu2+(aq) in the combined balanced equation?
When put together the balanced equation is just Cu + Cu+ = Cu2+ + Cu. So that means that the smallest possible integer coefficient for Cu2+ is just 1, right? If I'm wrong, can you please explain to me why I am wrong?
It should be 1, because the final balanced equation is 2Cu+ + Cu --> Cu2+ + 2Cu
Therefore, your answer is 1, because as you can see, the Cu^2+ has a coefficient of 1
this is because you combine the following half-reactions:
(1) Cu--> Cu 2+ + 2e-
(2) 2Cu + 2e- --> 2Cu
btw the second half-reaction has coefficients of 2 because I multiplied the entire equation by 2 in order to cancel out the 2 electrons on the product side of the first half reaction.
Is your equation balanced? You know it isn't because the charges don't balance.
So would it be Cu + 2Cu+ = Cu2+ + 2Cu? I didn't change the electrons before for the half reaction.
or cancel 1 Cu(s) for
2Cu^+ ==> Cu^2+ + Cu which is the disproportionation reaction for Cu^+. 1 goes down to Cu(s) and the other goes up to Cu^2+.
Well, you're almost there, but let me clarify it further for you. The half-reactions for the given galvanic cell are as follows:
Oxidation half-reaction: Cu(s) → Cu+(aq) + e^-
Reduction half-reaction: Cu2+(aq) + e^- → Cu(s)
Now, to form the balanced equation, we need to ensure that the electrons cancel out. So, if we multiply the oxidation reaction by 2 (to balance the electrons), we get:
2Cu(s) → 2Cu+(aq) + 2e^-
Cu2+(aq) + e^- → Cu(s)
Now, let's combine the two half-reactions by adding them:
2Cu(s) + Cu2+(aq) → 2Cu+(aq) + Cu(s)
So, the smallest possible integer coefficient for Cu2+(aq) in the combined balanced equation is 1.
In short, your initial understanding was correct. Keep up the good work, my punny friend!
Your understanding is incorrect. The combined balanced equation is not just Cu + Cu+ = Cu2+ + Cu.
To determine the balanced equation for the galvanic cell, we need to write the half-reactions for both the oxidation and reduction processes.
Oxidation Half-Reaction: Cu(s) → Cu2+(aq) + 2e-
Reduction Half-Reaction: Cu+(aq) + e- → Cu(s)
To balance the number of electrons, we need to multiply the reduction half-reaction by 2:
2Cu+(aq) + 2e- → 2Cu(s)
Now, we can write the overall balanced equation for the galvanic cell by combining the two half-reactions:
Cu(s) + 2Cu+(aq) → Cu2+(aq) + 2Cu(s)
In the balanced equation, the coefficient of Cu2+(aq) is 1, consistent with the stoichiometry obtained from the combined half-reactions.