Given the following two half reactions: Cd²⁺(aq) + 2e⁻ → Cd(s) E° = -0.40 V; Zr⁴⁺ (aq) + 4e⁻ → Zr(s) E° = -1.53 V. Determine E° and the spontaneity of the following reaction: 2 Cd²⁺(aq) + Zr(s) → 2 Cd(s) + Zr⁴⁺ (aq)
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To determine the standard cell potential (E°) of the reaction and its spontaneity, we need to add the half-reactions and their corresponding potentials.
The given half-reactions are:
1. Cd²⁺(aq) + 2e⁻ → Cd(s) with E°₁ = -0.40 V
2. Zr⁴⁺ (aq) + 4e⁻ → Zr(s) with E°₂ = -1.53 V
First, we need to reverse one of the half-reactions to match the electron transfer. Let's reverse the first half-reaction to balance the electrons:
Cd(s) + 2e⁻ → Cd²⁺(aq) with E°₁' = +0.40 V
Now, we can multiply each half-reaction by a suitable factor so that the electrons will cancel out when added together:
2(Cd²⁺(aq) + 2e⁻ → Cd(s)) → 2Cd²⁺(aq) + 4e⁻
Zr⁴⁺ (aq) + 4e⁻ → Zr(s)
Adding these two half-reactions:
2Cd²⁺(aq) + Zr⁴⁺ (aq) → 2Cd(s) + Zr(s)
Now, we can add the standard reduction potentials (E°) of the half-reactions to find the standard cell potential (E°) of the overall reaction:
E° = E°₁' + E°₂
E° = +0.40 V + (-1.53 V)
E° = -1.13 V
The standard cell potential (E°) of the reaction is -1.13 V.
To determine the spontaneity of the reaction, we can use the sign of the cell potential (E°):
- If E° > 0, the reaction is spontaneous (favorable).
- If E° < 0, the reaction is non-spontaneous (unfavorable).
In this case, since E° = -1.13 V (which is less than 0), the reaction is non-spontaneous (unfavorable) under standard conditions.