To find the voltage of the final cell formed by combining the two half cells, we can follow these steps:
### Step 1: Identify the half-reactions and their standard electrode potentials.
We have the following half-reactions:
1. Nickel half-reaction:
\[
Ni^{2+} + 2e^- \rightarrow Ni(s), \quad E^\circ = -0.25 \, V
\]
2. Zinc half-reaction:
\[
Zn^{2+} + 2e^- \rightarrow Zn(s), \quad E^\circ = -0.77 \, V
\]
### Step 2: Determine which half-reaction will undergo oxidation and which will undergo reduction.
- The half-reaction with the more positive standard electrode potential will be reduced, while the one with the more negative potential will be oxidized.
- Here, \( Ni^{2+} + 2e^- \) has a higher (less negative) potential than \( Zn^{2+} + 2e^- \).
### Step 3: Reverse the half-reactions as necessary.
- For zinc, we reverse the reaction for oxidation:
\[
Zn(s) \rightarrow Zn^{2+} + 2e^-, \quad E^\circ = +0.77 \, V
\]
- For nickel, we keep the reaction as is for reduction:
\[
Ni^{2+} + 2e^- \rightarrow Ni(s), \quad E^\circ = -0.25 \, V
\]
### Step 4: Calculate the standard cell potential.
The standard cell potential \( E^\circ_{cell} \) can be calculated using the formula:
\[
E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}
\]
Where:
- \( E^\circ_{cathode} \) is the reduction potential of nickel, which is \( -0.25 \, V \).
- \( E^\circ_{anode} \) is the oxidation potential of zinc, which is \( +0.77 \, V \).
Substituting the values:
\[
E^\circ_{cell} = (-0.25 \, V) - (+0.77 \, V) = -0.25 \, V - 0.77 \, V = -1.02 \, V
\]
### Step 5: Interpret the result.
The final voltage of the cell formed by combining the two half cells is \( -1.02 \, V \). However, since we are interested in the absolute value for the cell voltage, we can say that the cell voltage is \( 1.02 \, V \) when considering the direction of current flow.
### Final Answer:
The voltage of the final cell formed by combining the two half cells is \( 1.02 \, V \).
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