Questions 10-11 refer to the following information. [VISUAL] Standard reduction potentials for the h — Electrochemistry Chemistry Question

STEPS:

1. Identify the half-reactions and standard reduction potentials:
From the provided information, write down the reduction half-reactions and their associated standard reduction potentials ($E^\circ_{\text{red}}$):
* $\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ_{\text{red}} = +0.80\text{ V}$
* $\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad E^\circ_{\text{red}} = -0.76\text{ V}$

2. Determine which half-reaction undergoes oxidation and which undergoes reduction:
For a thermodynamically favorable reaction in a galvanic cell, the overall standard cell potential ($E^\circ_{\text{cell}}$) must be positive.
* The half-reaction with the more positive reduction potential ($\text{Ag}^+$) remains as a reduction at the cathode:
$$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ_{\text{red}} = +0.80\text{ V}$$
* The half-reaction with the more negative reduction potential ($\text{Zn}^{2+}$) must be reversed to represent oxidation at the anode:
$$\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \quad E^\circ_{\text{ox}} = +0.76\text{ V}$$
* Reversing this oxidation half-reaction yields a positive overall cell potential of $E^\circ_{\text{cell}} = +0.80\text{ V} - (-0.76\text{ V}) = +1.56\text{ V}$.

3. Balance the electrons transferred in the redox process:
In any balanced chemical equation, charge and mass must be conserved, meaning the number of electrons lost in oxidation must equal the number of electrons gained in reduction.
* The zinc oxidation half-reaction releases two electrons ($2e^-$).
* The silver reduction half-reaction accepts only one electron ($1e^-$).
* Multiply the silver reduction half-reaction by a factor of 2 to balance the electrons:
$$2\ \text{Ag}^+(aq) + 2e^- \rightarrow 2\ \text{Ag}(s)$$

4. Combine the half-reactions to generate the final net ionic equation:
Add the balanced reduction half-reaction and oxidation half-reaction together:
$$2\ \text{Ag}^+(aq) + \text{Zn}(s) + 2e^- \rightarrow 2\ \text{Ag}(s) + \text{Zn}^{2+}(aq) + 2e^-$$
Cancel the $2e^-$ from both sides of the equation to arrive at the final balanced net ionic equation:
$$2\ \text{Ag}^+(aq) + \text{Zn}(s) \rightarrow 2\ \text{Ag}(s) + \text{Zn}^{2+}(aq)$$
This matches Option D.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: It lists an impossible divalent silver cation, $\text{Ag}^{2+}(aq)$, which does not exist under these conditions and is not present in the given half-reactions.
• Option B is incorrect: It lists both $\text{Ag}^+(aq)$ and $\text{Zn}^{2+}(aq)$ as reactants. In the thermodynamically favorable cell reaction, $\text{Zn}^{2+}(aq)$ is a product resulting from the oxidation of solid zinc anode material, not a reactant.
• Option C is incorrect: This equation fails to balance both mass and charge. It does not scale up the silver species to account for the two electrons lost by zinc, and it leaves an extraneous free electron ($e^-$) on the product side. In a properly balanced net ionic equation, all transferred electrons must cancel out completely.