cell potential calculation

Question

For cell: Zn(s)|Zn²⁺(1 M)||Cu²⁺(1 M)|Cu(s), E°(Zn²⁺/Zn) = -0.76 V, E°(Cu²⁺/Cu) = +0.34 V. E°cell is:

RankGuru · Accepted Answer

1. **Identify Electrodes**: In the cell notation, the left side is always the **Anode** ($Zn$) and the right side is the **Cathode** ($Cu$).
2. **Formula**: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ (using reduction potentials).
3. **Values**:
   - $E^\circ_{cathode} = +0.34$ V.
   - $E^\circ_{anode} = -0.76$ V.
4. **Calculation**: $E^\circ_{cell} = (+0.34) - (-0.76) = 0.34 + 0.76 = +1.10$ V.
5. **Significance**: A positive $E^\circ_{cell}$ confirms that the cell reaction is spontaneous in the forward direction.

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cell potential calculation

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