Answer:
Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)
Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻
Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)
E°cell = 1.10 V
Explanation:
<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>
<em>Suppose we have the following half-reactions.</em>
<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V</em>
<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s) E°red = -0.76 V</em>
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To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).
Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V
Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻ E°red = -0.76 V
To get the overall equation we add both half-reactions.
Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)
The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E°cell = E°red, cat - E°red, an
E°cell = 0.34 V - (-0.76 V) = 1.10 V
Since E°cell > 0, the reaction is spontaneous.
