Electrochemistry - Galvanic Cells and Nernst Equation

Calculate the emf of the cell in which the following reaction takes place: $Ni(s) + 2Ag^{+}(0.002 M) \rightarrow Ni^{2+}(0.160 M) + 2Ag(s)$. Given that $E^{\circ}_{cell} = 1.05\text{ V}$.

1. Identify $n$: The number of electrons transferred is $n = 2$.
2. Apply Nernst Equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^2}$.
3. Substitute values: $E_{cell} = 1.05 - \frac{0.0591}{2} \log \frac{0.160}{(0.002)^2}$.
4. Calculate log term: $\frac{0.160}{0.000004} = 40,000$. $\log(40,000) = 4.602$.
5. Final calculation: $E_{cell} = 1.05 - (0.02955 \times 4.602) = 1.05 - 0.136 = 0.914\text{ V}$.

The Nernst equation is used here to find the potential under non-standard concentrations. Note that the concentration of $Ag^{+}$ is squared because its stoichiometric coefficient in the balanced equation is 2.

Calculate the standard Gibbs energy ($\Delta G^{\circ}$) for the reaction: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$. Given $E^{\circ}_{Zn^{2+}/Zn} = -0.76\text{ V}$, $E^{\circ}_{Cu^{2+}/Cu} = +0.34\text{ V}$, and $F = 96500\text{ C mol}^{-1}$.

1. Calculate $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 - (-0.76) = 1.10\text{ V}$.
2. Identify $n = 2$ (since $Zn \rightarrow Zn^{2+} + 2e^{-}$).
3. Use formula $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
4. $\Delta G^{\circ} = -(2) \times (96500) \times (1.10) = -212300\text{ J mol}^{-1} = -212.3\text{ kJ mol}^{-1}$.

The negative value of $\Delta G^{\circ}$ indicates that the reaction is thermodynamically spontaneous under standard conditions.

Calculate the emf of the cell represented below at $298\text{ K}$: $Mg(s) | Mg^{2+}(0.1\text{ M}) || Ag^{+}(0.0001\text{ M}) | Ag(s)$. Given $E^{\circ}_{Mg^{2+}/Mg} = -2.36\text{ V}$ and $E^{\circ}_{Ag^{+}/Ag} = 0.80\text{ V}$.

1. Determine $E^{\circ}_{cell}$: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$ $E^{\circ}_{cell} = 0.80\text{ V} - (-2.36\text{ V}) = 3.16\text{ V}$

2. Write the cell reaction: $Mg(s) + 2Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)$ Here, $n = 2$.

3. Apply Nernst Equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Ag^{+}]^2}$ $E_{cell} = 3.16 - 0.02955 \log \frac{0.1}{(10^{-4})^2}$ $E_{cell} = 3.16 - 0.02955 \log (10^{7})$ $E_{cell} = 3.16 - 0.02955 \times 7$ $E_{cell} = 3.16 - 0.20685 = 2.95315\text{ V}$

The standard cell potential is first calculated using reduction potentials. The Nernst equation is then used to account for non-standard concentrations, noting that the silver ion concentration is squared due to its stoichiometric coefficient.

A cell is constructed with a Chromium electrode in $Cr^{3+}$ solution and a Cadmium electrode in $Cd^{2+}$ solution. Determine the standard cell potential and identify the cathode. Given $E^{\circ}_{Cr^{3+}/Cr} = -0.74\text{ V}$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40\text{ V}$.

1. Identify Cathode and Anode: The electrode with the higher reduction potential acts as the cathode. Since $-0.40\text{ V} > -0.74\text{ V}$, the Cadmium electrode is the Cathode and Chromium is the Anode.

2. Calculate $E^{\circ}_{cell}$: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$ $E^{\circ}_{cell} = -0.40\text{ V} - (-0.74\text{ V})$ $E^{\circ}_{cell} = 0.34\text{ V}$

In a galvanic cell, the more positive (or less negative) reduction potential determines the cathode. The standard EMF is the difference between these potentials.