Calculate the EMF of the cell Ni Ni2+ (1.0m)| Au³ (1.0M) (Au. Given that E°= -0.25V for Ni2+ | Ni, E°= +1.5V for Au³+ | Au.

To calculate the electromotive force (EMF) of the cell, we can use the following standard formula:

\[
\text{EMF} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
\]

In this case:

- The reduction half-reaction at the cathode is \( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \), with a standard reduction potential \( E^\circ_{\text{cathode}} = +1.5 \, \text{V} \).
- The oxidation half-reaction at the anode is \( \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \), with a standard reduction potential \( E^\circ_{\text{anode}} = -0.25 \, \text{V} \).

Substituting these values into the formula:

\[
\text{EMF} = 1.5 \, \text{V} - (-0.25 \, \text{V}) = 1.5 \, \text{V} + 0.25 \, \text{V} = 1.75 \, \text{V}
\]

Thus, the EMF of the cell is **1.75 V**.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. How does the concentration of ions affect the EMF of the cell?
2. What is the Nernst equation, and how can it be applied to this cell?
3. How would the EMF change if the concentration of \( \text{Ni}^{2+} \) were different from 1.0 M?
4. What role does temperature play in the EMF of a galvanic cell?
5. How do you determine which metal acts as the anode and which as the cathode?

**Tip:** The standard electrode potential is always given for reduction reactions. When reversing a half-reaction (like oxidation), the sign of \( E^\circ \) is also reversed.

Learn how to calculate the electromotive force (EMF) of the Ni Ni2+ (1.0M) | Au³ (1.0M) cell using standard electrode potentials. This problem involves determining EMF using the formula EMF = E°cathode - E°anode, where E° values for Ni2+ | Ni and Au³+ | Au are given as -0.25V and +1.5V respectively. Suitable for advanced high school and college students studying electrochemistry.

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