Q. how to calculate e cell

What “E cell” usually means: the cell electromotive force, also called the cell potential, is the electrical potential difference between the two electrodes of an electrochemical cell. A positive cell potential indicates a spontaneous reaction under the given conditions.

Step 1. Identify the two half-reactions and their standard reduction potentials. Write each half-reaction as a reduction, and find its standard reduction potential \(E^\circ\) from a table. For example, common half-reactions are

\[ \text{Cu}^{2+} + 2\,\text{e}^- \; \text{becomes} \; \text{Cu}, \qquad E^\circ = +0.34\ \text{V} \]

\[ \text{Zn}^{2+} + 2\,\text{e}^- \; \text{becomes} \; \text{Zn}, \qquad E^\circ = -0.76\ \text{V} \]

Step 2. Decide which half-reaction is the cathode and which is the anode. The cathode is the half-reaction with the larger (more positive) \(E^\circ\) value and is where reduction occurs. The anode is the half-reaction with the smaller \(E^\circ\) value and is where oxidation occurs.

Step 3. Compute the standard cell potential \(E^\circ_{\text{cell}}\). Use the formula

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}} \]

Important note: do not multiply the potentials by stoichiometric coefficients. Potentials are intensive properties.

Example calculation for the Cu / Zn cell. Copper has the more positive \(E^\circ\), so copper is the cathode and zinc is the anode. Thus

\[ E^\circ_{\text{cell}} = (+0.34\ \text{V}) – (-0.76\ \text{V}) = +1.10\ \text{V} \]

Step 4. If the cell is not at standard conditions, use the Nernst equation to get the actual cell potential \(E_{\text{cell}}\). The general Nernst equation is

\[ E_{\text{cell}} = E^\circ_{\text{cell}} – \dfrac{RT}{nF}\,\ln Q \]

where \(R\) is the gas constant \(8.3145\ \text{J mol}^{-1}\text{K}^{-1}\), \(T\) is the temperature in kelvins, \(F\) is the Faraday constant \(96485\ \text{C mol}^{-1}\), \(n\) is the number of electrons transferred in the balanced cell reaction, and \(Q\) is the reaction quotient formed from the activities (or approximate concentrations) of products over reactants raised to their stoichiometric powers.

At \(25^\circ\text{C}\) (298.15 K), the factor \(\dfrac{RT}{F}\ln 10\) evaluates to approximately \(0.05916\ \text{V}\). Therefore a convenient base-10 form is

\[ E_{\text{cell}} = E^\circ_{\text{cell}} – \dfrac{0.05916}{n}\,\log_{10} Q \]

Step 5. Construct the reaction quotient \(Q\). Write the net cell reaction (with electrons cancelled) and form \(Q\) as

\[ Q = \dfrac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \]

Use activities for precise work, or concentrations for dilute solutions. Solids and pure liquids are omitted from \(Q\).

Example continuation, nonstandard conditions. For the Zn / Cu cell the net reaction is zinc metal plus aqueous copper(II) becomes zinc(II) plus copper metal. The reaction quotient is

\[ Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]

If, for example, \([\text{Zn}^{2+}] = 1.00\ \text{M}\) and \([\text{Cu}^{2+}] = 0.0100\ \text{M}\), and \(n = 2\), then

\[ Q = \dfrac{1.00}{0.0100} = 100 \]

Now apply the Nernst equation at 25 °C, with \(E^\circ_{\text{cell}} = 1.10\ \text{V}\):

\[ E_{\text{cell}} = 1.10\ \text{V} – \dfrac{0.05916}{2}\,\log_{10}(100) \]

\[ E_{\text{cell}} = 1.10\ \text{V} – 0.02958 \times 2 = 1.10\ \text{V} – 0.05916\ \text{V} = 1.04084\ \text{V} \]

Round as appropriate, for example \(E_{\text{cell}} \approx 1.04\ \text{V}\).

Summary checklist to calculate \(E_{\text{cell}}\):

1. Write both half-reactions as reductions and look up \(E^\circ\) for each. 2. Identify cathode (higher \(E^\circ\)) and anode (lower \(E^\circ\)). 3. Compute \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}\). 4. If nonstandard conditions, write the net reaction and form \(Q\), determine \(n\), and use the Nernst equation \(E_{\text{cell}} = E^\circ_{\text{cell}} – \dfrac{RT}{nF}\ln Q\), or at 25 °C use \(E_{\text{cell}} = E^\circ_{\text{cell}} – \dfrac{0.05916}{n}\log_{10} Q\).