Zn⁰(s) + Cu⁺²(aq) => Zn⁺²(aq) + Cu⁰(aq)
Oxidation => Zn(s) => Zn⁺²(aq) + 2e¯ (The loss of 2e¯ means Zn⁺²(aq) has moved into solution leaving 2e¯ behind in the Zn(s) electrode which are responsible for the anodic negative charge on the anode.
Reduction => Cu⁺²(aq) + 2e¯ => Cu⁰(s) (The gain of 2e¯ by the Cu⁺²(aq) in solution leaves the copper electrode deficient in electrons and is responsible for the cathodic positive charge on the cathode.
In the zinc-copper cell, Zn(s) | Zn^+2(1M) || Cu^+2(1M) | Cu(s), which electrode is negative?
Cu^+2
Cu(s)
Zn(s)***
Zn^+2
1 answer