Write down half-cell and overall cell reaction for the following cell. Calculate its cell potential (E_cell) at 25 °C. Zn(s)|Zn^2+(aq)||Ni^2+(aq)|Ni(s) (Given: E° (Zn^2+/Zn) = -0.7... Write down half-cell and overall cell reaction for the following cell. Calculate its cell potential (E_cell) at 25 °C. Zn(s)|Zn^2+(aq)||Ni^2+(aq)|Ni(s) (Given: E° (Zn^2+/Zn) = -0.76 V, E° (Ni^2+/Ni) = -0.25 V and [Zn^2+] = 0.1 M, [Ni^2+] = 0.01 M) Read more

Steps to Solve

1. Identify Half-Cell Reactions

The half-cell reactions for the cell are as follows:

• For the zinc electrode (anode): $$ \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2\text{e}^- $$

• For the nickel electrode (cathode): $$ \text{Ni}^{2+}(aq) + 2\text{e}^- \rightarrow \text{Ni(s)} $$

2. Calculate Standard Cell Potential (E°)

Using the given standard electrode potentials:

$$ E^{\circ}{cell} = E^{\circ}{cathode} - E^{\circ}_{anode} $$

Substituting in values:

$$ E^{\circ}_{cell} = (-0.25 \text{ V}) - (-0.76 \text{ V}) = 0.51 \text{ V} $$

3. Use the Nernst Equation

The Nernst equation is given by:

$$ E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln Q $$

At 25°C (298 K), this simplifies to:

$$ E_{cell} = E^{\circ}_{cell} - 0.0592/n \log Q $$

Where ( n = 2 ) (number of electrons transferred), and ( Q ) is the reaction quotient.

4. Calculate the Reaction Quotient (Q)

For the cell reaction:

$$ Q = \frac{[\text{Zn}^{2+}]}{[\text{Ni}^{2+}]} $$

Substituting the given values:

$$ Q = \frac{0.1}{0.01} = 10 $$

5. Substitute Q into the Nernst Equation

Now we substitute ( Q ) back into the Nernst equation:

$$ E_{cell} = 0.51 \text{ V} - \frac{0.0592}{2} \log 10 $$

6. Calculate the Cell Potential (E_cell)

Calculate the logarithm:

$$ \log 10 = 1 $$

Substituting this value:

$$ E_{cell} = 0.51 \text{ V} - \frac{0.0592}{2} \cdot 1 $$

Simplifying gives:

$$ E_{cell} = 0.51 \text{ V} - 0.0296 \text{ V} = 0.4804 \text{ V} $$