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

STEPS:

1. Identify the standard reduction potentials from the exam data:
According to the table provided in the prompt, the standard reduction potentials ($E^\circ_{\text{red}}$) for the two half-reactions are:
* Silver reduction: $\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ_{\text{red}} = +0.80\text{ V}$
* Zinc reduction: $\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad E^\circ_{\text{red}} = -0.76\text{ V}$

2. Determine the oxidation and reduction components of the cell:
As established in our balanced net ionic equation from the previous question, the thermodynamically favorable reaction in a galvanic cell must yield a positive overall cell potential ($E^\circ_{\text{cell}} > 0$).
* The species with the more positive standard reduction potential ($\text{Ag}^+$) undergoes reduction at the cathode:
$$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ_{\text{cathode}} = +0.80\text{ V}$$
* The species with the more negative standard reduction potential ($\text{Zn}$) must undergo oxidation at the anode, which requires reversing its reduction half-reaction:
$$\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \quad E^\circ_{\text{anode}} = -0.76\text{ V}$$

3. Calculate the standard cell potential ($E^\circ_{\text{cell}}$):
The overall cell potential is calculated using the formula:
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$
Substitute the values into the equation:
$$E^\circ_{\text{cell}} = +0.80\text{ V} - (-0.76\text{ V}) = \mathbf{1.56\text{ V}}$$

4. Apply the thermodynamic rules for intensive properties:
A critical concept tested here is that standard reduction potentials ($E^\circ$) are intensive properties; they depend on the chemical nature of the species, not on the amount of material. Even though the silver reduction half-reaction must be doubled to balance the electrons transferred in the net ionic equation ($2\text{Ag}^+ + \text{Zn} \rightarrow 2\text{Ag} + \text{Zn}^{2+}$), the standard reduction potential ($E^\circ$) of silver is not multiplied by 2. It remains exactly $+0.80\text{ V}$. This calculation identifies Option C as the correct choice.

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WHY_OTHERS_WRONG:

• Option A is incorrect (0.04 V): This value is obtained by simply adding the two reduction potentials directly from the table ($+0.80\text{ V} + (-0.76\text{ V}) = 0.04\text{ V}$) without reversing the sign of the zinc anode reaction to represent oxidation.
• Option B is incorrect (0.84 V): This represents the arbitrary addition of the cathode reduction potential ($+0.80\text{ V}$) to the incorrect sum of the two reduction potentials ($0.04\text{ V}$), which has no valid electrochemical basis.
• Option D is incorrect (2.36 V): This is a common student error where the student mistakenly doubles the reduction potential of the silver half-reaction because its stoichiometric coefficient in the balanced equation is 2 ($2 \times 0.80\text{ V} - (-0.76\text{ V}) = 1.60\text{ V} + 0.76\text{ V} = 2.36\text{ V}$). This violates the principle that cell potentials are intensive properties and do not scale with reaction coefficients.