Calculation of Standard Cell Potential
What is Standard Cell Potential?
Standard cell potential
The standard cell potential, denoted as $E^\circ_{\text{cell}}$, is the voltage produced by an electrochemical cell under standard conditions.
These conditions include:
- A temperature of 298 K (25°C),
- A pressure of 100 kPa for gases,
- A 1.0 M concentration for all aqueous solutions.
The cell potential depends on the difference in the ability of two half-cells to gain or lose electrons, as measured by their standard electrode potentials ($E^\circ$).
The Formula for Standard Cell Potential
The standard cell potential is calculated using the formula:
$$
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
$$
Here’s what each term means:
- $E^\circ_{\text{cathode}}$: The standard electrode potential of the reduction half-equation occurring at the cathode.
- $E^\circ_{\text{anode}}$: The standard electrode potential of the oxidation half-equation occurring at the anode.
- Always subtract the (E^\circ) value of the anode (where oxidation occurs) from that of the cathode (where reduction occurs).
- Remember: reduction happens at the cathode, and oxidation happens at the anode.
Predicting Spontaneity with $E^\circ_{\text{cell}}$
The sign of $E^\circ_{\text{cell}}$ tells us whether the reaction in the electrochemical cell is spontaneous:
- If $E^\circ_{\text{cell}} > 0$: The reaction is spontaneous, meaning it can generate electrical energy.
- If $E^\circ_{\text{cell}}< 0$: The reaction is non-spontaneous, meaning it requires an external energy source to proceed.
- Do not confuse the signs of $E^\circ_{\text{cell}}$.
- A positive value indicates a spontaneous reaction, not a negative one.
Step-by-Step: Calculating $E^\circ_{\text{cell}}$
Let’s break down the process of calculating the standard cell potential with an example.
Zinc-Copper Electrochemical Cell
You’re tasked with calculating $E^\circ_{\text{cell}}$ for a cell consisting of a zinc electrode and a copper electrode.
The half-reactions and their standard electrode potentials are:
$$
\text{Zn}^{2+} (\text{aq}) + 2\text{e}^- \rightarrow \text{Zn} (\text{s}) \quad E^\circ = -0.76 \, \text{V}
$$
$$
\text{Cu}^{2+} (\text{aq}) + 2\text{e}^- \rightarrow \text{Cu} (\text{s}) \quad E^\circ = +0.34 \, \text{V}
$$
Step 1: Identify the Cathode and Anode
- The cathode is where reduction occurs.
- Since copper has a more positive $E^\circ$, it will be reduced: $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$.
- The anode is where oxidation occurs.
- Zinc has a more negative $E^\circ$, so it will be oxidized: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-$.
Step 2: Apply the Formula
$$
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
$$
Substitute the values:
$$
E^\circ_{\text{cell}} = (+0.34 \, \text{V}) - (-0.76 \, \text{V})
$$
Simplify:
$$
E^\circ_{\text{cell}} = +1.10 \, \text{V}
$$
Step 3: Interpret the Result
- Since $E^\circ_{\text{cell}} = +1.10 \, \text{V}$, the reaction is spontaneous.
- This means the zinc-copper cell can generate electricity.
- In the zinc-copper cell, zinc is oxidized to $\text{Zn}^{2+}$, releasing electrons that flow through an external circuit to reduce $\text{Cu}^{2+}$ to copper metal.
- This flow of electrons is what powers devices.
- Do not multiply $E^\circ$ values by coefficients when balancing half-equations.
- The potential is an intensive property and does not depend on the amount of substance.
Calculate $E^\circ_{\text{cell}}$ for a cell comprising a magnesium electrode and a silver electrode. Use the following data:
- $\text{Mg}^{2+} (\text{aq}) + 2\text{e}^- \rightarrow \text{Mg} (\text{s}) \quad E^\circ = -2.37 \, \text{V}$
- $\text{Ag}^{+} (\text{aq}) + \text{e}^- \rightarrow \text{Ag} (\text{s}) \quad E^\circ = +0.80 \, \text{V}$
Solution
- Identify the cathode and anode:
- Silver ($E^\circ = +0.80 \, \text{V}$) is the cathode (reduction).
- Magnesium ($E^\circ = -2.37 \, \text{V}$) is the anode (oxidation).
- Apply the formula:
$$
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
$$
$$
E^\circ_{\text{cell}} = (+0.80 , \text{V}) - (-2.37 \, \text{V}) = +3.17 \, \text{V}
$$ - Interpret the result: $E^\circ_{\text{cell}} = +3.17 \, \text{V}$ indicates a highly spontaneous reaction.
What is the significance of a positive $E^\circ_{\text{cell}}$?
