For the galvanic cell: `Zn(s) | Zn^(2+)(aq) (1.0 M) || Ni^(2+)(aq) (1.0 M) | Ni(s)`, `E_cell^o` will be [Given `E_((Zn^(2+)) /(Zn))^0 = -0.76 V`, `[E_((Ni^(2+))/(Ni))^0= -0.25V]`

Options

A. `-0.51V`
B. `-1.01V`
C. `0.51V`
D. `1.01V`

Answer

To find the standard cell potential \( E_{cell}^o \) for the given galvanic cell, we will follow these steps:

### Step 1: Identify the half-reactions
In the given galvanic cell, we have:
- Zinc (Zn) is oxidized to zinc ions (Zn²⁺).
- Nickel ions (Ni²⁺) are reduced to nickel (Ni).

The half-reactions can be written as:
1. Oxidation half-reaction (anode):
\[ \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \]
2. Reduction half-reaction (cathode):
\[ \text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \]

### Step 2: Write the standard reduction potentials
From the problem statement, we have the following standard reduction potentials:
- For zinc:
\[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V} \]
- For nickel:
\[ E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.25 \, \text{V} \]

### Step 3: Convert the oxidation potential for zinc
Since zinc is being oxidized, we need to convert the reduction potential to oxidation potential:
\[ E^\circ_{\text{oxidation}} = -E^\circ_{\text{reduction}} = +0.76 \, \text{V} \]

### Step 4: Calculate the standard cell potential
The standard cell potential \( E_{cell}^o \) can be calculated using the formula:
\[ E_{cell}^o = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E_{cell}^o = E^\circ_{\text{Ni}^{2+}/\text{Ni}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} \]
\[ E_{cell}^o = (-0.25 \, \text{V}) - (-0.76 \, \text{V}) \]
\[ E_{cell}^o = -0.25 \, \text{V} + 0.76 \, \text{V} \]
\[ E_{cell}^o = 0.51 \, \text{V} \]

### Final Answer
Thus, the standard cell potential \( E_{cell}^o \) is \( 0.51 \, \text{V} \).

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